Hey MHF i need some help with this proof.

If\(\displaystyle f(g)= t^m\) and \(\displaystyle g(t)= t^n\), where m and n are positive integers, show that:

\(\displaystyle (f*g)(t)= t^{m+n+1} \int_0^1 \! u^m(1-u)^n \, du.\)

show : \(\displaystyle \int_0^1 \! u^m(1-u)^n \, du = \frac{m!n!}{(m+n+1)!}\)

Use integration by parts \(\displaystyle n \) times ,

\(\displaystyle \int_0^1 \! u^m(1-u)^n \, du \)

\(\displaystyle = \left[ \frac{ u^{m+1} }{m+1} (1-u)^n \right]_0^1 + \frac{n}{m+1} \int_0^1 u^{m+1} (1-u)^{n-1}~du\)

\(\displaystyle = \frac{n}{m+1} \int_0^1 u^{m+1} (1-u)^{n-1}~du \)

\(\displaystyle = \frac{n(n-1)}{(m+1)(m+2)} \int_0^1 u^{m+2} (1-u)^{n-2}~du\)

\(\displaystyle = ... = \frac{n(n-1)...1}{(m+1)(m+2)(m+3)...(m+n)} \int_0^1 u^{m+n}~du \)

\(\displaystyle = \frac{n!}{(m+1)(m+2)(m+3)...(m+n)(m+n+1)} \)

\(\displaystyle = \frac{m! n!}{m! (m+1)(m+2)(m+3)...(m+n)(m+n+1)} \)

\(\displaystyle = \frac{m! n!}{(m+n+1)!} \)