Proposition:
$$ P_n \ : \sqrt[n]{n!} \le \frac{n+1}{2} , \hspace{0.5em}\forall n \in \mathbb{Z}^+ $$
Proof:
$$ P_1 : \text{Let} \ n=1 , \hspace{1em}\sqrt{1!} \le \frac{1+1}{2} \implies 1 = 1 , \ \therefore P_1 \text{is true} $$
$$ P_k : \text{Suppose} \ \exists k \ \text{s.t.} \hspace{1em}\sqrt[k]{k!} \le \frac{k+1}{2} , \hspace{0.5em}\forall k \in \mathbb{Z}^+ $$
$$ P_{k+1} : \text{Consider now ,} \hspace{1em} \sqrt[k+1]{(k+1)!} \le \frac{k+2}{2} $$
$$ \implies \sqrt[k+1]{(k+1) \cdot k \cdot (k-1)... 2\cdot1} \le \frac{k+2}{2} $$
Question:
I am unsure how to bring $P_k$ into this expression and proceed further. What should I consider to solve this problem?