Prove: Γ(ν +1)·(ν + 1)(ν + 2)(ν + 3) ...... (ν + k) is equal to Γ(ν + k +1).

Proof. The gamma function can be viewed as a generalized factorial function and, indeed is often called that. The argument of the gamma function is a real number while the factorial function is defined only for positive integers. However for the special case when the argument of the gamma function is an integer n, the gamma function is related to the factorial function by

            Γ(n) = (n - 1)!

The recursion formula for the factorial function is

            n!(n+1) = (n+1)!

while the recursion formula for the gamma function is

1)        νΓ(ν) = Γ(ν +1) .

Let us apply the recursion formula 1) to the product

2)        Γ(ν +1)·(ν + 1)(ν + 2)(ν + 3) ...... (ν + k) .

Using 1) we infer that Γ(ν +1)·(ν + 1) = Γ(ν + 2) so replacing Γ(ν +1)·(ν + 1) in 2) with Γ(ν + 2) gives

3)        Γ(ν + 2)·(ν + 2)(ν + 3) ...... (ν + k) .

Now replacing Γ(ν + 2)·(ν + 2) in 3) with Γ(ν + 3) gives

3)        Γ(ν + 3)·(ν + 3) ...... (ν + k) .

Repeating this process we end up replacing Γ(ν + k)·(ν + k) with Γ(ν + k + 1). Thus we have proved that

            Γ(ν +1)·(ν + 1)(ν + 2)(ν + 3) ...... (ν + k) = Γ(ν + k + 1)