<s>[INST] Prove by induction that $ \ \forall n \ge 2, \ 2^{n+1} > n^{2} + 3$ Question:

Prove by induction that $$ \ \forall n \ge 2, \  2^{n+1} > n^{2} + 3$$

My attempt:
Base case is trivial.
Suppose that $ n \ge 2$ and   $\  2^{n+1} > n^{2} + 3$
WTS $ \  2^{n+2} > (n+1)^{2} + 3$
$ 2^{n+2} = 2.2^{n+1}> 2(n^2 + 3) = 2n^2 + 6 = n^2 + n^2 + 2n -2n +6 = (n^2+2n+1) + 3 + (n^2 -2n+2) = (n+1)^2 + 3 + (n^2 -2n+2)$.
I am not sure what to do from here. How can I show $ (n^2 -2n+2) > 0?$
 [/INST] $n^2-2n+2=(n-1)^2+1>0$.
By the way, this reasoning gives also the proof without an induction:
$$2^{n+1}-n^2-3=(1+1)^{n+1}-n^2-3\geq$$
$$\geq1+(n+1)+\frac{(n+1)n}{2}+\frac{(n+1)n(n-1)}{6}-n^2-3=$$
$$=\frac{(n-1)(n^2-2n+6)}{6}>0.$$
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