Prove $\left(1-\frac{1}{n^2}\right)^n\times\left(1+\frac 1n\right)<1$ for every natural $n > 0$ by induction Consider this inequality 
$$\left(1-\dfrac{1}{n^2}\right)^n\times\left(1+\dfrac 1n\right)<1$$ 
which is meant to be valid for any nonzero natural number $n$.
It is asked to prove it by induction. I haven't made any significant progress after many tries.
Any advice is welcome.
Note :
A way I encountered which looks nicer and more straightforward is to notice that 
$1-\dfrac{1}{n^2}<1-\dfrac{1}{n^2+k}$ for $k\in \{1,\cdots,n\}$
Multiplying side by side those $n$ inequalities, a nice telescoping takes place to give exaclty the sought result.

The inequality 
$\left(1-\dfrac{1}{n^2}\right)^n\cdot\left(1+\dfrac 1n\right)
=\left(1-\dfrac{1}{n}\right)^n\left(1+\dfrac{1}{n}\right)^{n+1}
=\dfrac{\left(1+\dfrac{1}{n}\right)^{n+1}}{\left(1+\dfrac{1}{n-1}\right)^{n}}<1$
is equivalent to the sequence 
$\left\{\left(1+\dfrac{1}{n}\right)^{n+1}\right\}$
is strictly decreasing. 
So we will show that：sequence 
$\left\{\left(1+\dfrac{1}{n}\right)^{n+1}\right\}$
is strictly decreasing.
By AG-MG inequality
$\dfrac{1}{\left(1+\dfrac{1}{n}\right)^{n+1}}=\left(\dfrac{n}{n+1}\right)^{n+1}
=1\cdot\dfrac{n}{n+1}\cdot\dfrac{n}{n+1}\cdots\dfrac{n}{n+1}
<\left(\dfrac{n+1}{n+2}\right)^{n+2}$
$=\left(\dfrac{n+1}{n+2}\right)^{n+2}=\dfrac{1}{\left(1+\dfrac{1}{n+1}\right)^{n+2}}.$
Hence $$\left(1+\dfrac{1}{n}\right)^{n+1}>\left(1+\dfrac{1}{n+1}\right)^{n+2}.$$
