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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!$

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Best way is to try a small example $2\cdot 4 \cdot 6=48=2^3\cdot 6$ and not $2\cdot 6$ - that way you also get more of a feel for what is going on. – Mark Bennet May 17 '14 at 20:01


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

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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 $n!$

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