How to prove $n^{3/2} \le 2^n/n$ for all $n\geq 8$ via Induction? How to prove $n^{3/2} \le 2^n/n$ for all $n \geq 8$
My attempt:
$\textbf{Base case}$ is $n_0=8$: $\quad   \left(8^{3/2}=4\right)\leq \left(32=\frac{2^8}{8}\right) \checkmark$
$\textbf{Induction hypothesis}$: $\exists n\in \mathbb{N}_{\geq 8}:(n+1)^{3/2}\leq \frac{2^{n+1}}{n+1}$
$\textbf{Inductive step}$: 
$$
\begin{gather}
(n+1)^{3/2}\leq \frac{2^{n+1}}{n+1} \end{gather} \\
  \iff (n+1)^{5/3}\leq 2^{n+1} \\
  \iff (n+1)^{5/3}\leq 2^n \cdot 2 \quad |\uparrow ^{3/5} \\
  \iff n+1\leq 2^{3n/5}\cdot 2^{3/5}\ $$
This, sadly, doesn't get me anywhere, any ideas?

Continuing Sonnhard's answer. Hint: $$ (2n^{5/2})^{2} = 4 n^{5} \ge n^{5} + n^{5} + n^{5} + n^{5} \ge  n^{5} + 8 n^{4} + 64 n^{3} + 8^{3} n^{2}  $$
Or can also 
$$n^{5} + n^{5} + n^{5} + n^{5} \ge  n^{5} + 16 n^{4} + 64 n^{3}   $$

$$ (n+1)^{5} = n^{5} + \binom{5}{1}n^{4} + \binom{5}{2} n^{3} + \binom{5}{4} n^{2} +1$$
