Prove that $(n+1)^{n-1}<n^n$

For $n\gt1$, Bernoulli's Inequality gives $$ \begin{align} \frac{n^n}{(n+1)^{n-1}} &=n\left(1-\frac1{n+1}\right)^{n-1}\\ &\ge n\left(1-\frac{n-1}{n+1}\right)\\ &=\frac{2n}{n+1}\\[6pt] &\gt1 \end{align} $$

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We can also use Bernoulli's Inequality slightly differently. If $n\gt1$, then Bernoulli's Inequality is strict since $\frac1{n+1}\ne0$: $$ \begin{align} \frac{n^n}{(n+1)^{n-1}} &=(n+1)\left(1-\frac1{n+1}\right)^n\\ &\gt(n+1)\left(1-\frac{n}{n+1}\right)\\[6pt] &=1 \end{align} $$

Divide both sides of $(n+1)^{n-1}<n^n$ by $n^{n-1}$. We have now $$\frac{(1+1/n)^n}{1+1/n}<n.$$ The left side tends to $e$ from below. $e$ is less than 3 and the inequality holds for $n=2$.

For $n > 1$, apply GM $\le$ AM to following $n$ numbers $\big(\;\underbrace{n+1,n+1,\ldots,n+1}_{n-1 \text{ copies}}, 1\;\big)$. We have

$$((n+1)^{n-1}\cdot 1)^{1/n} \le \frac{(n-1)(n+1)+1}{n} = n \quad\implies\quad (n+1)^{n-1} \le n^n$$ Since the $n$ numbers are not the same, the inequality is strict and we are done.