proving by multiplying Let (*) be $ \left(1+\frac{1}{n}\right)\left(1-\frac{1}{n^2}\right)^n < 1, ∀n∈ℕ$  
I have tried many ways to get to (*)
We have $\left(1-\frac{1}{n^2}\right)^n <1$ and $-\left(1+\frac{1}{n}\right) < -1 <1$  
How can you infer (*) from these two last inequalities?  

Proof: We know that $a_n=(1+\frac{x}{n})^n$ is a strictly increasing sequence for $x\geq -1$ and it is bounded above to $e^x$.Letting $x=\frac{-1}{n}$, we get: $$(1-\frac{1}{n^2})^n < e^{-1/n}$$
Therefore,
$$(1+\frac{1}{n})(1-\frac{1}{n^2})^n < (1+\frac{1}{n})\times e^{-1/n}$$
But, it is a well-known fact that for all real values $x \in \mathbb{R}$, we have $e^x \geq 1+x$ (a proof is available here). Hence,
$$e^{1/n}\geq 1+\frac{1}{n} \implies (1+\frac{1}{n})\times e^{-1/n} \leq 1$$
And we obtain:
$$(1+\frac{1}{n})(1-\frac{1}{n^2})^n < 1$$
Q.E.D.
Addendum:
To know why $a_n=(1+\frac{x}{n})^n$ is bounded above to $e^x$, note that $a_n$ is strictly increasing and $\lim_{n\to \infty}a_n=e^x$. So, it has to be bounded from the above by $e^x$. To know why $a_n$ is strictly increasing, from the AM-GM inequality we have:
$$\sqrt[n+1]{x_{1}x_{2}\cdots x_{n+1}}<\frac{x_{1}+x_{2}+\ldots+x_{n+1}}{n+1}%$$
Let $x_1=1$ and $x_2=\cdots=x_{n+1}=1+\frac{x}{n}$
$$
\left(  1+\frac{x}{n}\right)  ^{\frac{n}{n+1}}<\frac{1+n\left(  1+\frac{x}{n}\right)  }{n+1}=1+\frac{x}{n+1}%
$$
$$a_n=\left(  1+\frac{x}{n}\right)^{n}<\left(1+\frac{x}{n+1}\right)^{n+1}=a_{n+1}
$$
Remark:
It's irrelevant but interesting to see that using the Bernoulli's inequality we get a lower bound for the expression as follows:
$$(1-\frac{1}{n^2})^n \geq 1-\frac{1}{n}$$
$$(1+1/n)(1-\frac{1}{n^2})^n \geq \frac{n+1}{n}\frac{n-1}{n}=\frac{n^2-1}{n^2}$$
Therefore, 
$$\frac{n^2-1}{n^2}<(1+\frac{1}{n})(1-\frac{1}{n^2})^n$$
Therefore, as you could see, proving that it's smaller than $1$ is actually harder than I initially thought because the expression quickly gets very close to $1$. ;-)
