Proving that $\forall n\in \mathbb N, 2^n |(2n)!$

Question

I'm having some trouble proving the following statement:

$$\forall n\in \mathbb N, \ \ 2^n |(2n)!$$

The first thing that came to mind was mathematical induction but this is my first time using it with this kind of the question (with divisibility instead of an equality/inequality) and I'm not quite sure how to proceed. How can I prove this? Should I use the properties of the divisibility of two numbers and manipulate $2^n |(2n)!$ until I get $2^{n+1} |(2(n+1))!$ or should I use the fact that $\exists k \in \mathbb Z: (2n)! = k2^n $ and try to conclude that $\exists l\in \mathbb Z:\ (2(n+1)) = l 2^{n+1}!$ ?

There are $n$ even numbers amongst $1,2, \cdots, 2n$. And that doesn't count the multiples of $4, 8, $ and so on. — lulu, Dec 21 '20 at 15:22
The proposition is true for $n=1$. Assume, for some $k$, $2^k \mid (2k)!$. Then, $(2(k+1))! = (2k)!(2k+1)(2k+2)$. Hence, $2^{k+1} \mid (2(k+1))!$. — Aaratrick, Dec 21 '20 at 15:24
Induction is a good approach here. Start with the basis case, $n=1$. — hardmath, Dec 21 '20 at 15:24
That's not really an induction question: just list all even factors in $(2n)!$. There're $n$ of them. QED. — GNUSupporter 8964民主女神地下教會, Dec 21 '20 at 15:26

score 1 · Accepted Answer · answered Dec 21 '20 at 15:27

Induction works best.

For $n=1$, $2^1 =2$ divides $2!=2$.

Suppose $2^n$ divides $(2n)!$. Then consider $2^{n+1}=2\cdot 2^n$. By induction hypthesis, $2^n$ divides $(2n)!$ and $(2(n+1))! =(2n)! \cdot [((2n)!+1)\cdot ((2n)^!+2)\cdots (2(n+1))!]$. The latter partial product $[\ldots]$ contains obviously a factor which is divisible by $2$. I.e., $2^{n+1}$ divides $(2(n+1))!$ as claimed.

Proving that $\forall n\in \mathbb N, 2^n |(2n)!$

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