Standard inductive problem Question: Prove that $2^n \geq (n+1)^2$ for all $n \geq 6$.
I have tried to prove this below and I'm interested if my method was correct and if there is a simpler answer since my answer seems unnecessarily long for such a simple claim.
Inductive hypothesis $$2^n \geq (n+1)^2$$ 
We need to show that $2^{n+1} \geq (n+2)^2$ or alternatively
$$2^n2 \geq (n+1)^2 \frac{(n+2)^2}{(n+1)^2}$$
claim $$\frac{(n+2)^2}{(n+1)^2} <2, \forall n \geq 6$$ 
notice that it is true for $n=6$ and $$\frac{\frac{(n+2)^2}{(n+1)^2}}{\frac{((n+1)+2)^2}{((n+1)+1)^2}}<1$$ so $\frac{(n+2)^2}{(n+1)^2}$ is decreasing as $n$ gets larger so we have proven the above claim.
So we have that for all $n \geq 6$ we have $2^n \geq (n+1)^2$ and by our induction hypothesis and $\frac{(n+2)^2}{(n+1)^2} <2$ so we can conclude that $2^n2 \geq (n+1)^2 \frac{(n+2)^2}{(n+1)^2}$.

You have $2^{n+1}=2\times 2^{n}\geq 2(n+1)^2$
Now, for $n\geq 6,$ 
$$2(n+1)^2-(n+2)^2$$
$$=2(n^2+2n+1)-(n^2+4+4n)$$
$$=2n^2+4n+2-n^2-4-4n$$
$$=n^2-2\geq0$$
$$\therefore 2(n+1)^2\geq(n+2)^2$$
$$\therefore 2^{n+1}\geq(n+2)^2$$ 
:)
