Use induction to prove that that $8^{n} | (4n)!$ for all positive integers $n$ Use induction to prove that that $8^{n} | (4n)!$ for all positive integers $n$
So far I have:
Base case (n = 1)
= $8^{1} | (4(1))!$
= $8 | 24$ which is true.
Induction Step:
$8^{n + 1} | (4(n + 1))!$
$8^{n + 1} | (4n + 4)!$


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*A bit confused as to how to close this proof out, also wanted to make sure my current progress is correct as well. Any help is appreciated.


Well, you know that $(4n)! = 8^nk$ for some integer $k$.
$$\Rightarrow (4n + 4)! = (4n+4)(4n+3)(4n+2)(4n+1)(4n)!$$
$$ = 8^nk(4n+4)(4n+3)(4n+2)(4n+1)$$
Factor out of the even terms:
$$ = 8^n 8 k (n+1)(4n+3)(2n+1)(4n+1)$$
$$= 8^{n+1} k'$$
For some integer $k'$. This means that $(4n+1)!$ is divisible by $8^{n+1}$.
