Suppose that a function $f$ is analytic inside and on a positively oriented circle $C_R$, centered at $z_0$ and with radius $R$. Then by Cauchy Integral Formula
$$f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_{C_R}\frac{f(z)}{(z-z_0)^{n+1}}dz.$$
Since $C_R$ is compact (i.e. closed and bounded) and $f(z)$ is continuous, $|f(z)|$ assumes a maximum (and also a minimum) on $C_R$. Let us call the maximum value of $|f(z)|$ on $C_R$ $N_R$. On $C_R$, $|z-z_0|=R$. Hence, we have the upper bound for $|f^{(n)}(z_0)|$ as
\begin{align*}
|f^{(n)}(z_0)|&=\frac{n!}{2\pi}\left|\oint_{C_R}\frac{f(z)}{(z-z_0)^{n+1}}dz\right|\\
&\leq\frac{n!M_R}{R^n}
\end{align*}
for $n=1,2,\cdots$. This inequality is called Cauchy’s Inequality.

Using Cauchy’s Inequality, we can prove the following Liouville’s Theorem.

Theorem. If $f$ is entire and bounded in the complex plane $\mathbb{C}$, then $f(z)$ is constant.

Proof. Since $f$ is bounded, there exists $M>0$ such that $|f(z)|\leq M$ for all $z\in\mathbb{C}$. Since $f$ is entire, for any $R>0$,
$$|f'(z)|\leq\frac{M}{R}$$
by Cauchy’s Inequality. As $R\to\infty$, $\frac{M}{R}\to 0$, so we obtain $|f'(z)|=0$ for all $z\in\mathbb{C}$. This implies that $f'(z)=0$ for all $z\in\mathbb{C}$ and hence, $f(z)$ is constant.

Cauchy’s Integral Theorem says that if a function $f(z)$ is analytic throughout some simply connected domain $D$, then for any contour $C$ in $D$, $\oint_C f(z)dz=0$. It turns out that the converse of Cauchy’s Theorem is also true, namely

Theorem. If a function $f(z)$ is continuous in a simply connected domain $D$ and $\oint_C f(z)dz=0$ for every closed contour $C$ within $D$, then $f(z)$ is analytic throughout $D$.

This theorem is called Morera’s Theorem. Let us prove the theorem.

First we show that under the assumption the line integral $\int_\gamma f(z)dz$ is independent on the path $\gamma$. Let $\gamma$ be a path from $z_1$ to $z_2$ and choose $\gamma_1$ another path from $z_1$ to $z_2$ as shown in the figure.

Then $C: \gamma\cup (-\gamma_1)$ is a positively oriented contour. So, by assumption,
\begin{align*}
0&=\oint_C f(z)dz\\
&=\int_\gamma f(z)dz+\int_{-\gamma_1}f(z)dz\\
&=\int_\gamma f(z)dz-\int_{\gamma_1}f(z)dz.
\end{align*}
That is,
$$\int_\gamma f(z)dz=\int_{\gamma_1}f(z)dz.$$
Hence, the line integral $\int_\gamma f(z)dz$ does not depend on the path $\gamma$ but only on the endpoints $z_1$ and $z_2$.

Let us define $F(z)=\int_{z_0}^z f(w)dw$. Then
\begin{align*}
F(z+\Delta z)-F(z)&=\int_{z_0}^{z+\Delta z} f(w)dw-\int_{z_0}^z f(w)dw\\
&=\int_z^{z+\Delta z} f(w)dw.
\end{align*}
Since $\int_z^{z+\Delta z}dw=\Delta z$,
$$f(z)=\frac{1}{\Delta z}\int_z^{z+\Delta z}f(z) dw$$
and so, we have
$$\frac{F(z+\Delta z)-F(z)}{\Delta z}-f(z)=\frac{1}{\Delta z}\int_z^{z+\Delta z}[f(w)-f(z)]dw.$$
Since $f$ is continuous at $z$, given $\epsilon>0$ there exists $\delta>0$ such that $|f(w)-f(z)|<\epsilon$ whenever $|w-z|<\delta$. If $z+\Delta z$ is close enough to $z$ so that $|\Delta z|<\delta$, then
$$\left|\frac{F(z+\Delta z)-F(z)}{\Delta z}-f(z)\right|<\frac{1}{|\Delta z|}\epsilon|\Delta z|=\epsilon.$$
Therefore,
$$F'(z)=\lim_{\Delta z\to 0}\frac{F(z+\Delta z)-F(z)}{\Delta z}=f(z).$$
That is, $F(z)$ analytic in $D$. This means that $f(z)$ is analytic by Cauchy Integral Formula.

Suppose that $f(z)$ is analytic everywhere inside and on a simple closed contour $C$, taken in the positive sense. If $z_0$ is a point exterior to $C$, then by Cauchy’s Integral Theorem,
$$\oint_C\frac{f(z)}{z-z_0}dz=0.$$
Now, the question is what would be the value of the integral
$$\oint_C\frac{f(z)}{z-z_0}dz$$
if $z_0$ is a point interior to $C$? To answer this question, let us consider a tiny circle centerd at $z_0$ and with radius $r$, also oriented counterclockwise as shown in the figure.

Then using Cauchy’s Integral Theorem, we obatin
$$\oint_C\frac{f(z)}{z-z_0}dz-\oint_{C_1}\frac{f(z)}{z-z_0}dz=0.$$
Let $z=z_0+re^{i\theta}$. Then
\begin{align*}
\oint_{C_1}\frac{f(z)}{z-z_0}dz&=\int_0^{2\pi}\frac{f(z_0+re^{i\theta})}{re^{i\theta}}rie^{i\theta}d\theta\\
&=i\int_0^{2\pi}f(z_0+re^{i\theta})d\theta\\
&\to 2\pi if(z_0)
\end{align*}
as $r\to 0$. Therefore, we obtain
$$f(z_0)=\frac{1}{2\pi}\oint_C\frac{f(z)}{z-z_0}dz.$$
This formula is called Cauchy Integral Formula.

Example. Let $C$ be the positively oriented circle $|z|=2$. The function $f(z)=\frac{z}{9-z^2}$ is analytic within and on $C$. Since $z_0=-i$ is interior to $C$, by Cauchy Integral Formula,
\begin{align*}
\oint_C\frac{z}{(9-z^2)(z+i)}dz&=\oint_C\frac{\frac{z}{9-z^2}}{z+i}dz\\
&=2\pi\frac{-i}{9-(-i)^2}\\
&=\frac{\pi}{5}.
\end{align*}

Derivatives

Suppose that $|\Delta z_0|$ is small enough so that $z_0+\Delta z_0$ is still interior to the contour $C$. Then by Cauchy Integral Formula, we obtain
$$\frac{f(z_0+\Delta z_0)-f(z_0)}{\Delta z_0}=\frac{1}{2\pi i\Delta z_0}\left\{\oint_C\frac{f(z)}{z-z_0-\Delta z_0}dz-\oint_C\frac{f(z)}{z-z_0}dz\right\}.$$
So,
\begin{align*}
f'(z_0)&=\lim_{\Delta z_0\to 0}\frac{f(z_0+\Delta z_0)-f(z_0)}{\Delta z_0}\\
&=\lim_{\Delta z_0\to 0}\frac{1}{2\pi i\Delta z_0}\left\{\oint_C\frac{f(z)}{z-z_0-\Delta z_0}dz-\oint_C\frac{f(z)}{z-z_0}dz\right\}\\
&=\lim_{\Delta z_0\to 0}\frac{1}{2\pi i\Delta z_0}\oint_C\frac{\Delta z_0f(z)}{(z-z_0-\Delta z_0)(z-z_0)}dz\\
&=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^2}dz.
\end{align*}
The same process can be repeated for $f'(z_0)$ and we obtain
$$f^{\prime\prime}(z_0)=\frac{2}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^3}dz.$$
Continuing, we get
$$f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}$$
for $n=0,1,2,\cdots$ where $f^{(0)}(z)=f(z)$. This formula may also be called Cauchy Integral Formula which includes the Cauchy Integral Formula we derived earlier.

Example. If $C$ is the positively oriented unit circle $|z|=1$ and $f(z)=\exp(2z)$, then
\begin{align*}
\oint_C\frac{\exp(2z)}{z^4}dz&=\oint_C\frac{f(z)}{(z-0)^{3+1}}\\
&=\frac{2\pi i}{3!}f^{\prime\prime\prime}(0)\\
&=\frac{8\pi i}{3}
\end{align*}
where $f(z)=\exp(2z)$.

Example. Let $z_0$ be any point interior to a positively oriented simple closed contour $C$. When $f(z)=1$, Cauchy Integral Formula tells that
$$\oint_C\frac{dz}{z-z_0}=2\pi i\ \mbox{and}\ \oint_C\frac{dz}{(z-z_0)^{n+1}}=0,\ n=1,2,\cdots.$$