I have to prove the inequality $n! \geq (\frac{n}{e})^n$ by using the Cauchy estimate on $f(z) = e^z$.
I found a proof that goes something like this:
I know that by using the Cauchy integral formula $f^n(a) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z-a)^{n+1}}dz$, for $a = 0$, I have $f^n(0) = \frac{n!}{2\pi i} \oint_\gamma \frac{e^z}{z^{n+1}}dz = 1$ for every derivative, because the derivative of $e^z$ is always $e^z$ and evaluated at $z= 0$ is $1$.
The Cauchy estimate is $|f^n(a)| \leq Mn!r^{-n}$ where $|f(z)| \leq M$ on $\gamma$. If I have $|z| = r$ for $\gamma$, I get $|f(z)| = e^r \leq M$. So for the Cauchy estimate I get $\frac{1}{n!} \leq \frac{e^r}{r^n}$. Now, here it says that by using calculus, $\frac{e^r}{r^n}$ is minimized at $r=n$. By putting this in the inequality and rearranging one gets the desired result.
My question is, how does this step with $\frac{e^r}{r^n}$ minimized at $r=n$ works?