Solve by induction: $n!>(n/e)^n$

Here's a hint. You assume that $n!>\left(\frac{n}{e}\right)^n$. Now you should show it for n+1, i.e., you should show that $(n+1)! > \left(\frac{n+1}{e}\right)^{n+1}$.

You can write

\begin{equation} (n+1)! > (n+1) \left(\frac{n}{e}\right)^n = (n+1)\left(\frac{n}{n+1}\right)^n \left(\frac{(n+1)^n}{e^n}\right) \end{equation}

Can you solve it from here?

By considering the exponential power series we observe that for $x>0$, $$ e^x > \frac{x^n}{n!} $$ Now setting $x=n$ we obtain $$e^n > \frac{n^n}{n!} $$ which rearranges to precisely what is desired. I should note that I had first learned this incredibly short and simple proof of this fact from Qiaochu Yuan's posts on this website, and he in turn attributed it to this article written by Terence Tao.

This is the "easy" part of Stirling's formula -- the crude order of magnitude estimate showing how huge $n!$ is, without the $\sqrt{2 \pi n}$ correction that is harder to derive.

Taking logarithms, you are asking for $\log(n!) > n (\log n - 1)$. The latter is the indefinite integral of $\log(n)$. Drawing a picture, this follows from $\log(n)$ being an increasing function. The inequality compares the area under the graph of the function to the area of rectangles underneath the graph. A stronger inequality can be obtained using trapezoids and the convexity of $\log(x)$. I think you can get the $\sqrt{n}$ factor this way but not the exact constant $\sqrt{2 \pi}$.