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.