<s>[INST] 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 $2^n2 \geq (n+1)^2 \frac{(n+2)^2}{(n+1)^2}$.
 [/INST] Except for showing a base case I think your method is good.  Note that without a base case you do not have an induction, so make sure you pick e.g., $n=6$ and note that $2^6=64\ge (6+1)^2=49.$  You might consider trying to show the inductive hypothesis by showing that $n\ge 6\implies (n+1)^2\ge 2n+3$ and thus 
$$2\cdot 2^n\ge 2(n+1)^2 \ge (n+1)^2+2n+3=(n+2)^2$$
where $2n+3$ is the difference between $(n+1)^2$ and $(n+2)^2$.
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