<s>[INST] Proof by induction, $2^{n} > n^{2} - 2$ So I have to proof that $2^{n} > n^{2} - 2$ for $n > 2$, using mathematical induction.
I start of with the "basecase" for n = 3:
RHS: $2^{3} = 8$, LHS: $3^{2} - 2 = 7$
Clearly, the inequality is correct in this case. I assume that the inequality is true for $\forall n = k$ where $k \in \mathbb{Z}^{+}$.
To prove this inequality, we have to proof that it holds for n = k + 1.
$2^{k} > k^{2} - 2$
We multiply both sides with 2:
$2^{k+1} > 2(k^{2} - 2)$
$2^{k+1} > 2k^{2} - 4$
$2^{k+1} > k^{2} + k^{2} - 2 - 2$
According to our assumption,  $2^{k} > k^{2} - 2 $, hence $2^{k+1} > k^{2} - 2 $
Because $k^{2} - 2 \geq 2k + 1 $ for $k > 2 $, we get:
$2^{k+1} > k^{2} + 2k + 1 - 2$
Thus:
$2^{k+1} > (k+1)^{2} - 2$
QED.
How does my proof look, does it have any deficits, and how would you prove it?
Thank you!
 [/INST] I think this is a little clearer.
$2^{k+1}\ge2k^2-4$
$=k^2+2k+1-2+k^2-2k-3$
$=(k+1)^2-2+(k-3)(k+1)$
$\ge(k+1)^2-2$ for $k\ge3$
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