and then factor out 2 from each even number. For
example,

`4*3*2*1 = (4*2)*(3*1) = 2^2*(2*1)*(3*1)`

Direct answer to the
question is as follows:

The statement in the question is true for
`n=0`.

Since `2^n` is an even number for `n ge 1`,

`2^n! =
2^n*(2^n-1)*(2^n-2)*(2^n-3)*(2^n-4)cdots4*3*2*1`

`=
{2^n*(2^n-2)*(2^n-4)cdots4*2}*{(2^n-1)*(2^n-3)cdots3*1}`

`=
2^(2^(n-1))*{2^(n-1)*(2^(n-1)-1)*(2^(n-1)-2)*(2^(n-1)-3)cdots4*3*2*1}*k_n`

,w
here `k_n = (2^n-1)*(2^n-3)cdots3*1` and it is an odd number.
Therefore,

`2^n! = 2^(2^(n-1))*(2^(n-1)!)*k_n`

Applying the same procedure
to the above `2^(n-1)!` until there are no even numbers,

`=
2^(2^(n-1))*2^(2^(n-2))*2^(2^(n-3))cdots2^(2^2)*2^(2^0)*(2^0!)*K_n`

,where
`K_n = k_n*k_(n-1)cdotsk_1` and it is an odd number.

`=
(2^(2^(n-1)+2^(n-2)+2^(n-3)+cdots+2^1+2^0))*K_n`

Since
`1+2+4+cdots+2^(n-2)+2^(n-1) = 2^n-1`,

`= 2^(2^n-1)*K_n`

Therefore,
`2^(2^n-1)` divides `2^n!` for `n ge 0`.