The function $f (n) = (1 + 1 / n) ^ {n+1}$ is decreasing

I cannot prove that the function $$f (n) = \left(1 + \frac1n\right) ^ {n + 1},$$ defined for every positive integer $n$, is strictly decreasing in $n$. I already tried to prove by induction and also tried to prove by calculating the difference between $f (n + 1)$ and $f (n)$. I need help.

May be this help:

$f_n/f_{n+1}=\frac{(1+1/n)^{n+1}}{(1+1/(n+1))^{n+2}}=(\frac{1+1/n}{1+1/(n+1)})^{n+1}\times\frac{1}{1+1/(n+1)}=(1+1/(n^2+2n))^{n+1}\times\frac{1}{1+1/(n+1)}$

But

$$(1+1/(n^2+2n))^{n+1}>1+\frac{n+1}{n^2+2n}>1+\frac{n+1}{(n+1)^2}=1+1/(n+1)$$

$$f(n)=\exp{\left((n+1)\ln{\left(1+\frac1n\right)}\right)}$$ with $f(n)>0$. Hence $$f'(n)=f(n)\left(\ln{\left(1+\frac1n\right)-\frac1n}\right)$$ Now, use the inequality $\ln(1+x)<x$ for $x>-1$ with $x=\frac1n$ to conclude that $f'(n)<0$.

Take the fraction $$ \frac{f(n)}{f(n-1)}=\left(\frac{n+1}{n}\right)^{n+1}\bigg/\left(\frac{n}{n-1}\right)^n=\left(\frac{(n-1)(n+1)}{n^2}\right)^n\frac{n+1}{n}=\frac1{\left(1+\frac{1}{n^2-1}\right)^n}\frac{n+1}{n}. $$ Using the induction, show that for $n\ge2$ and $x>-1$, $x\ne0$ (if $x=0$ then there would be a trivial equality) $(1+x)^n>1+nx$. For $n=2$ we have $(1+x)^2=1+2x+x^2>1+2x$. For $n+1$: $(1+x)^{n+1}=(1+x)^n(1+x)>(1+nx)(1+x)=1+(n+1)x+nx^2>1+(n+1)x$.

Using this inequality one can conclude that $$ \left(1+\frac{1}{n^2-1}\right)^n>1+n\frac1{n^2-1}=\frac{n^2+n-1}{n^2-1}, $$ and $$ \frac{f(n)}{f(n-1)}=\frac1{\left(1+\frac{1}{n^2-1}\right)^n}\frac{n+1}{n}<\frac{n^2-1}{n^2+n-1}\frac{n+1}{n}=\frac{n^3+n^2-n-1}{n^3+n^2-n}<1. $$ Therefore, $f(n)$ is decreasing in $n$.