Proof that: $n! = \Omega (5^{\log n})$

Question

Show that: $n! = \Omega (5^{\log n})$

That is to find a $c$ and a $n_0$ such that: $$n! \geq c\cdot5^{\log n}$$

We have:

$$n!\leq n^n$$

Thus,

$$n\log n \geq \log n \log 5 + \log c$$

Which is: $$(n - \log 5)\log n \geq \log c$$

So we can pick $c = 1$ and $n_0 = 3$, and it works. Is that right?

Answer 1

\begin{align}n!=\Omega(5^{\log_2n})&\impliedby n!\ge c\cdot5^{\log_2n}=c\cdot n^{\log_25}\quad\forall n\ge n_0\\&\impliedby n(n-1)(n-2)\ge n^{\log_25}\quad (c=1)\\&\impliedby n^{3-\log_25}-\frac3{n^{\log_25-2}}+\frac2{n^{\log_25-1}}\ge1\quad(\log_25\in[2,3])\\&\text{true with}\,n_0=5\end{align}