<s>[INST] Inequality proof by induction of $n!>2^n+3^n$ Give a proof by induction: 
$\forall n\ge n_0 \in \mathbb{N}$ we have $n!>2^n+3^n$.
Attempt:
Base case: $n_0=7$
Induction step:
$k!>2^k+3^k$
Prove $(k+1)!>3^{k+1}+2^{k+1}$
We know $(k+1)!>(k+1)(2^k+3^k)$
as $(k+1)>6 \because k\ge 7$:
$(k+1)!>6(2^k+3^k)$
$(k+1)!>3*2^{k+1}+2*3^{k+1}$
as $3*2^{k+1}+2*3^{k+1}>3*2^{k+1}+2*3^{k+1}-2*2^{k+1}-3^{k+1}$
$(k+1)!>2^{k+1}+3^{k+1}$
still getting my head around this method of induction with inequalities.
Any help is greatly appreciated, thank you!
 [/INST] From here for $k> 2$ we have
$$(k+1)!\stackrel{I.H.}>(k+1)(2^k+3^k)=(k+1)2^k+(k+1)3^k>2\cdot 2^k+3\cdot 3^k=2^{k+1}+3^{k+1}$$
and this prove the induction step.
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