How to verify $(1+\frac{1}{n})^2(1-\frac{1}{n^2})^{n-1}\geq \exp(\frac{1}{n})$ How to verify this inequality? Assuming that $n\in \mathbb{N}^+$, we have:
$$\left(1+\frac{1}{n}\right)^2\left(1-\frac{1}{n^2}\right)^{n-1}\geq \exp\left(\frac{1}{n}\right).$$

Consider $$A_n=\left(1+\frac{1}{n}\right)^2\times\left(1-\frac{1}{n^2}\right)^{n-1}$$ Take logarithms $$\log(A_n)=2\log\left(1+\frac{1}{n}\right)+(n-1)\log\left(1-\frac{1}{n^2}\right)$$ Now, use the Taylor series $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right)$$ and replace $x$ by $\frac 1n$ in the first term and by $-\frac 1{n^2}$ in the second term. 
This should give $$\log(A_n)=\frac{1}{n}+\frac{1}{6 n^3}+\frac{1}{15
   n^5}+O\left(\frac{1}{n^6}\right) >\frac 1n$$
You can check, using limits, that $A_1=4 > e$.
Edit
Taking more terms, the expansion of $\log(A_n)$ does not seem to contain any negative coefficient (I did stop at $O\left(\frac{1}{n^{1000}}\right)$, using,  for sure, a CAS for this last part).
A closer look at the coefficients reveals that $$\log(A_n)=\sum_{k=1}^\infty \frac{1}{k (2 k-1)}\frac 1 {n^{2 k-1}}$$
