<s>[INST] Proof by induction for $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$ for $k > 4$ I was given this proof for hw. 
Prove that $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$
So, far I've gotten this
Basis:
$k = 5$, $2^{5 + 1} - 1 > 2\cdot5^2 + 2\cdot5 + 1$ => $63 > 61$ (So, the basis holds true)
Hypothesis:
for all $k > 4$, $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$
Inductive step:
LHS 
$2^{k + 1 + 1} - 1   
= 2^{k + 2}   
= 2 \cdot 2^{k + 1}$
$2[ 2^{k + 1} - 1 + 1] - 1 \geqslant 2\cdot(2k^2 + 2k + 1 + 1) - 1$
RHS (Attempt to prove that the RHS is less than the LHS )
$2(2k^2 + 2k + 1 + 1) - 1 \geqslant 2(k + 1)^2 + 2(k + 1) + 1$
Next step
$2(k + 1)^2 + 2(k + 1) + 1   
= 2k^2 + 4k + 2 + 2k + 1 + 1   
= (2k^2 + 2k + 1) + 4k + 3$
Now I'm stuck at how to prove that
$2(2k^2 + 2k + 1 + 1) - 1 \geqslant (2k^2 + 2k + 1) + 4k + 3$
So, what can I do?
Thanks
 [/INST] Inductive step
$$2^{k+2}-1=2\times 2^{k+1}-1>2(2k^2+2k+2)-1>2(k+1)^2+2(k+1)+1$$
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