Help with proof that $n! \geqslant e(\frac{n}{e})^n$

Question

For all $n \in \mathbb{Z}_+$ show that $$n! \geqslant e \left( \frac{n}{e} \right)^n. $$

I tried this myself.

BS: For $n=1$ we have $1 \geqslant 1$.

IH: Let $\mathcal{P}(k)$ be the statement that $k! \geqslant e \left( \frac{k}{e} \right)^k$.

IS: In proving $\mathcal{P}(k+1)$ I get $$(k+1)! = (k+1)k! \geqslant (k+1)e \left( \frac{k}{e} \right)^k, $$ but how do I prove $$ (k+1) \left( \frac{k}{e} \right)^k \geqslant \left( \frac{k+1}{e} \right)^{k+1}? $$

Answer 1

This is equivalent to $$e\ge\left(1+\frac1k\right)^k.$$ To prove this, you could prove that $\log(1+x)\le x$ for all $x\ge1$.