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

How would one prove that $$(n+1)^{n-1}<n^n \ \forall n>1$$

I have tried several methods such as induction.

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

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.

Proof By Induction: The hypothesis is true for $n=2$. Let it be true for $n=k$. Then $$(k+1)^{k-1}<k^k\\\implies \left(1+\frac{1}{k}\right)^{k}<{k+1}$$Now, $$\frac{(k+2)^k}{(k+1)^{k+1}}=\left(1+\frac{1}{k+1}\right)^{k}\frac{1}{k+1}<\left(1+\frac{1}{k}\right)^{k}\frac{1}{k+1}<1$$

Proof By Calculus: Let us define the function $f:\mathbb{R}^+\to \mathbb{R},\ f(x)=(x-1)\ln (x+1)-x\ln x$. Then, since $f(1)=0$, we need to show that $f$ is decreasing on $[1,\infty)$. Note that $f'(x)=\ln (x+1)+\frac{x-1}{x+1}-\ln x-1=\ln (x+1)-\ln x-2/(x+1)\implies f''(x)=\frac{1}{x+1}-\frac{1}{x}+\frac{4}{(x+1)^2}=\frac{x(x+1)-(x+1)^2+4x}{x(x+1)^2}=\frac{3x-1}{x(x+1)}>0\ \forall x\in [1,\infty)\implies f'(x)$ is increasing in $[1,\infty)\implies f'(x)<\lim_{x\to \infty}f'(x)=0\ \forall x\in [1,\infty)\implies f$ is decreasing in $[1,\infty)$.

My try to prove $\forall n>1 : (n+1)^{n-1} < n^n$:

$$(n+1)^{n-1} < n^n \Leftrightarrow (n+1)^{n} < n^n (n+1) \Leftrightarrow \left(\frac{n+1}{n}\right)^n < n+1$$

Now apply induction on this result:

Base case: $n = 2 \Rightarrow \left(\frac{3}{2}\right)^2 < 3 \Rightarrow 2.25 < 3$ OK

Induction hypothesis: assume $\left(\frac{n+1}{n}\right)^n < n+1$ true for all $n \leq k$

Induction step: we prove that $\left(\frac{n+1}{n}\right)^n < n+1$ is true for $n = k+1$ $$\left(\frac{k+2}{k+1}\right)^{k+1} < k+2 \Leftrightarrow \left(\frac{k+2}{k+1}\right)^k < k+1$$ Now we have that $\left(\frac{k+1}{k}\right)^k > \left(\frac{k+2}{k+1}\right)^k$ and because of the induction hypothesis, we have proven by induction that $\forall n>1 : \left(\frac{n+1}{n}\right)^n < n+1$ and thus that $\forall n>1 : (n+1)^{n-1} < n^n$