Bernoulli's Inequality: using estimate from $a>0$ to show the estimate for $n\geq2$

Question

I'm trying to use the estimate $(1+a)^n\geq1+na$ for $a>0$ to show for all $n\geq2$ the estimate $(1+\frac{1}{n^2 -1})^n>1+\frac{1}{n}$

$(1+a)^n\geq1+na+\frac{n(n-1)}{2}a^2$

$(1+n)^0=1$

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

Answer 1

We have $(1+\frac{1}{n^2 -1})^n \ge 1+\frac{n}{n^2-1}$

From $n^2 >n^2-1$ we get: $\frac{n}{n^2-1}>\frac{1}{n}$

Hence $(1+\frac{1}{n^2 -1})^n>1+\frac{1}{n}$