<s>[INST] Proof by induction $\left(1 - \frac{1}{n^{2}}\right)^n \left(1 + \frac{1}{n}\right) < 1$ I'm trying to prove that for every integer $n > 0\quad (P)$ : 
$$\left(1 - \frac{1}{n^{2}}\right)^n \left(1 + \frac{1}{n}\right) < 1$$
This is what I did so far, using induction : 
$$\text{for }(n = 1) : (1 - 1/1^2)^1 (1 + 1/1) = 0 < 1$$
Let's suppose that $(P)$ is true for all integers from $1$ to a certain $n$ 
and let's prove that $P$ is true for $n+1$ also : 
I simplified the formula and now it looks like this : 
$$(P) : \left(\frac{n-1}{n}\right)^n < \left(\frac{n}{n+1}\right)^{(n+1)}$$
So if we define $Q(n) = (\frac{n-1}{n})^n$
We have $Q(n) < Q(n+1)$
and we need to prove that $Q(n+1) < Q(n+2)$
That's where my head stopped. Can anyone help me please ? 
 [/INST] Consider the function 
\begin{eqnarray*}
y= \frac{x^{x+1}}{(1+x)^{x+1}} \\
\frac{1}{y} \frac{dy}{dx} =\frac{1}{x} +\ln x - \ln (1+x) \color{yellow}{>0} 
\end{eqnarray*}
Now for $x>0$ we have $e^{\frac{1}{x}} > 1+ \frac{1}{x}$ & so 
\begin{eqnarray*}
x+1 < x e^{\frac{1}{x}} \\
\ln (x+1) < \frac{1}{x}+ \ln(x).  
\end{eqnarray*}
In other words, our function $y$ is monotonically increasing & so
\begin{eqnarray*}
\frac{(n-1)^n}{n^n} < \frac{n^{n+1}}{(1+n)^{1+n}} 
\end{eqnarray*}
This can be rearranged to give your result.
</s>