I'd appreciate if somebody could check whether my proof (mathematical induction) is correct.
$$(\forall n \in \mathbb{N}, n\geq 5) 2^{n+3} < (n+1)!$$
(i) let $n=5$ $2^{5+3} = 2^{8} < (5+1)!=6!$
(ii) Let's assume that the statement is true for $k \in \mathbb{N}, k\geq 5$.
Our inductive hypothesis (IH) is that: $2^{k+3} < (k+1)!$
We'll show that: $2^{k+4} < (k+2)!$
$$2^{k+4} = 2.2^{k+3} < (IH) 2.(k+1)!$$
since $k\geq 5 \Rightarrow k+2\geq 7$ then...
$$ 2.(k+1)! < (k+2)(k+1)! = (k+2)!$$
By principle of mathemathical induction, we've shown that the statement applies for $\forall n\in \mathbb{N}, n\geq 5$.
Is the second part correct? Im worried whether it's okay for me to use the fact that $k\geq 5 \Rightarrow k+2\geq 7$ then...
$$ 2.(k+1)! < (k+2)(k+1)! = (k+2)!$$