Prove that $e^n n! \geq n^n$.

Question

I want to prove that \begin{equation}\label{eq1} e^n n! \geq n^n,\: \forall \: n \in \mathbb{N^*}. \end{equation}

We will proceed by induction. For n=1 is true.

Suppose valid for some $n \in \mathbb{N^*}$.

We will prove that it is valid for $n+1$. Equivalently, we will prove that $$ (n+1)+\ln((n+1)!) \geq (n+1) \ln(n+1).$$ Indeed,

\begin{eqnarray} (n+1)+\ln((n+1)!)& = & (n+1)+\ln((n+1)n!)\\ & = & n+1+\ln(n+1)+\ln n! \\ & \geq & n \ln n+1+\ln(n+1). \end{eqnarray}

I thought to write $ \ln n = \int_{1} ^ {n} \frac{1}{x}\;dx $.But I couldn't get out of that last line. How to continue? Or is that not the way?

Answer 1

You don't have to use induction; one can observe that

$e^n = \sum_{j=0}^{\infty}\frac{n^j}{j!}$

Thus $e^n n! = \sum_{j=0}^{\infty}\frac{n^jn!}{j!} \geq \frac{n^nn!}{n!} = n^n$

Answer 2

A much simpler rearrangement with induction:

$$\frac{(n+1)^{n+1}}{(n+1)!}\frac{n!}{n^n}=\left(1+\frac1n\right)^n\le e\implies\frac{n^n}{n!}\le e^n$$