An identity for the product of even numbers (double factorial)

Question

I'm unable to prove this identity:

Prove that: $2\cdot 4 \cdot 6 \cdot 8 \cdots 2n = 2^n \cdot n!$

Wouldnt it be like this? $ 2(1 \cdot 2\cdot 3\cdot 4 \cdots n)= 2 \cdot n!$

Answer 1

$$\prod_{r=1}^n(2r)=2^n\prod_{r=1}^nr$$

as there are $n$ number of $2$s in the Left Hand Side as multiplier

Answer 2

No, Ashish, you factor out a 2 from each number. It is not the destributive property you are thinking of. You factor $n$ 2's hence the term $2^n$ in front of $n!$