Prove that n! is greater than n at a given power

Question

Can someone help me with this demonstration?

$n! \geq n^{log_2 5} $

for $n > 10$

Answer 1

Hint: Prove that $n! \ge n^3$ for $n \ge 10$. (It actually holds for $n\ge 6$).

Indeed, $$ n! = n(n-1)(n-2)\cdots 2 \cdot 1 > 2n(n-1)(n-2) > n^3 \mbox{ if } n\ge 6 $$

Answer 2

The ratio of two successive factorials is the growing factor $n$, and that of two powers is $(n/(n-1))^p$, which is bounded by the constant $(10/9)^p$ (for $n>10$). Sooner or later the LHS will exceed the RHS.