I want to prove, using mathematical induction, the following proposition:
$$ \forall n\in \:\mathbb{N}\:,\:\:\:\left(n+1\right)!\:\ge \:2^n $$
My thesis is this:
$$ \forall n\in \:\mathbb{N}\:,\:\:\:\left(p+2\right)!\:\ge \:2^{p+1} $$
Thank you for the help!
For the case $n=0$ you have that $1!=1\leq 2^0$, which is true. Now, if you asume that $(n+1)!\geq 2^n$ then $$(n+2)!=(n+1)!(n+2)\geq (n+2)2^n.$$ Since $n+2\geq 2$ for all $n \in \mathbb{N}$ then $$(n+2)!=(n+1)!(n+2)\geq (n+2)2^n\geq 2 \cdot 2^n=2^{n+1}.$$ By induction it follow that the statement is true for all natural $n$.