Showing $(n+1)^n<e^nn!$ by induction

Question

Show $(n+1)^n<e^nn!$

I know why that would be the case using general knowledge and a bit of substitution but am clueless on how to prove it.

Answer 1

Hint: Consider the simple factorization. \begin{equation} (n+1)^{n} = n^{n} \left( 1 + \frac{1}{n} \right)^{n}. \end{equation}

Answer 2

• The inequality is true for $n = 1$ because $(1+1)^1 = 2$ and $e^1 1! = e > 2$.

• Suppose it is true for some fixed $n$. Multiplying both sides by $(n+1)$, we obtain $$ (n+1)^{n+1} < e^n (n+1)! $$ Finally, notice that $(n+2)^{n+1}=(n+1)^{n+1}\left(1+\frac{1}{n+1}\right)^{n+1} < (n+1)^{n+1}\cdot e^1$, so that $$ (n+2)^{n+1} < e^{n+1}(n+1)! $$