Is the following proposition true or false? Justify your conclusion.
For each non-negative integer $n$, $8^n\mid(4n)!$.
Attempt:
I've tried to expand $(4(n+1))!$ to show that it's equivalent to $(4n+4)(4n+3)(4n+2)(4n+1)(4n)!$ and then I tried factoring these all, to try and show that they were all divisible by $8$. But I got lost because it seemed like I did it wrong, mainly because the $8^n$ was still there and unaccounted for.
I tried to prove that the claim $P(n)$ was true for $P(0), P(1), P(2)$, and $P(3)$.
Check that it holds for $n=1$.
Try mathematical induction suppose $$8^n|(4n)!$$
We can write $(4n)!=8^nm$, where $m\in \mathbb{Z}$.
We want to prove that $$8^{n+1}|(4n+4)!$$
\begin{align} (4n+4)! &= (4n)![(4n+4)(4n+3)(4n+2)(4n+1) ]\\ &= (8^n m)[\color{blue}4 (n+1)(4n+3)(\color{blue}2)(2n+1)(4n+1)] \end{align}
Can you continue from here?