How can I show that for a $n \in \mathbb N$

$$e^{1-n} \leq \frac {n!}{n^n}$$

I tried using the binomial theorem like this

$$n^n \le (1+n)^n = \sum_{k=0}^n \binom nk n^k \le \sum_{k=0}^\infty \binom nk n^k = \sum_{k=0}^\infty \frac{n!}{k!(n-k)!} n^k \le \sum_{k=0}^\infty \frac{n!}{k!} n^k = n! \sum_{k=0}^\infty \frac{n^k}{k!} = n! \cdot e^n$$

which would give me

$$\frac{1}{e^n} \le \frac{n!}{n^n}$$

But I'm missing the factor of $e$ on the left side. Can you give me a hint?

  • $\begingroup$ Have you tried applying induction? $\endgroup$
    – Mark Viola
    Jan 6 '18 at 22:43
  • $\begingroup$ Looks like Stirling's formula $\endgroup$
    – Dunham
    Jan 6 '18 at 22:48

\begin{align*} n^{n}=(1+(n-1))^{n}\leq n!\sum_{k=0}^{\infty}\dfrac{(n-1)^{k}}{k!}=n!\cdot e^{n-1}. \end{align*}


Using induction we see that for $n=1$, the inequality holds. Assume that it holds for some number $k$.

Then, using $\left(1+\frac1k\right)^k<e$, we find that

$$\begin{align} \frac{(k+1)!}{(k+1)^{k+1}}&=\frac{k!}{k^k\left(1+\frac1k\right)^k}\\\\ &\ge \frac{e^{1-k}}{e}\\\\ &=e^{1-(k+1)} \end{align}$$

And we are done!


Hint: $$ \frac{\frac{(n+1)^{n+1}}{(n+1)!}}{\frac{n^n}{n!}}=\left(1+\frac1n\right)^n\lt e $$


Note that

$$e^{1-n}\leq \frac {n!}{n^n}\iff \frac {e^nn!}{n^n}\geq e$$

that is true $\forall n$ indeed

$$n=1 \implies \frac{e^11!}{1^1}\geq e$$


$$\frac{a_{n+1}}{a_n}=\frac {e^{n+1}(n+1)!}{(n+1)^{n+1}}\frac {n^n}{e^nn!}=\frac{e}{\left(1+\frac1n\right)^n}>1$$


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