RTP : $$(1-\frac{1}{n^2})^n>1-\frac{1}{n}, n≥2,n∈N$$
Attempt at solution : Lets show the base case is true, i.e $n=2$. We do algebra and show that it is $\frac{9}{16}$>$\frac{1}{2}$. Next assume the expression is true when $n=k$, where $k≥2$, and prove it's true when $n=k+1$ $$(1-\frac{1}{k^2})^k>1-\frac{1}{k}$$ So we need to manipulate this expression, such that it becomes identical to the case where $n=k+1$, or, alternatively, such that the $LHS$ is identical with $k+1$ case. Then we could show it's greater than something which is greater than our $RHS$ etc. How is it possible to form the $k+1$ case when we have $k^2$ in the denumerator of our $LHS$? And If we can't do that, what other way is there to solve this problem?
HINT:
Given that you have to show an inequality, you do not need to work out the expressions exactly. For example, you can point out things like:
$$k+1>k$$
and
$$\frac{1}{k^2}>\frac{1}{(k+1)^2}$$
and
$$(1-\frac{1}{k^2})^k > (1-\frac{1}{k^2})^{k+1}$$
to get to the inequality you want.