Prove that $I_{m, n} = I_{m+1,n-1}$ for $n\geq 1$. 
Let $I_{m,n} = \dfrac{1}{m!n!}\displaystyle\int_0^1 x^m(1-x)^ndx$ for $m,n\in\mathbb{Z},m,n \geq 0$. Prove that $I_{m,n}=I_{m+1,n-1}$ for $n\geq 1$.

I thought of proving this by induction, with $P(n)$ being the statement that $I_{m,n} = I_{m+1,n-1}.$ I can prove the base case, but I'm having difficulty proving the inductive step. I tried using the hypothesis (i.e., $\dfrac{1}{m!k!}\displaystyle\int_0^1 x^m (1-x)^kdx=\dfrac{1}{(m+1)!(k-1)!}\displaystyle\int_0^1 x^{m+1}(1-x)^{k-1}dx$). But I'm not sure how to prove $I_{m,k+1} = I_{m+1,k}$ using integration by parts.
 A: Replacing $m$ by $m+1$ in $x^m$ is antidifferentiating (modulo multiplication by a constant).
Replacing $n$ by $n-1$ in $(1-x)^n$ is differentiating (modulo multiplication by a constant).
Differentiating one factor and antidifferentiating the other is exactly what happens in integration by parts.
So integrate by parts:
\begin{align}
& \int_0^1 x^m(1-x)^n\,dx \\[10pt]
= {} & \int_0^1 (1-x)^n \Big( x^m \, dx\Big) \\[10pt]
= {} & \int u \,dv = uv - \int v\,du \\[10pt]
= {} & \left[ (1-x)^n \frac{x^{m+1}} {m+1} \right]_0^1 - \int_0^1 \frac{x^{m+1}} {m+1} \cdot n(1-x)^{n-1}(-1) \, dx \\[10pt]
= {} & \frac n {m+1} \int_0^1 x^{m+1} (1-x)^{n-1} \, dx.
\end{align}
A: Just integrate $I_{m,n}$ by parts. Taking $u=(1-x)^n$ and $dv=x^mdx$ the integration by parts formula
$$\int_0^1u\cdot dv =\left[u\cdot v\right]_0^1-\int_0^1v\cdot du$$
gives
$$I_{m,n}={1\over m!n!}\left[(1-x)^n{x^m\over m+1}\right]_0^1+{1\over (m+1)!(n-1)!}\int_0^1x^{m+1}(1-x)^{n-1}dx$$
A: There is no need for induction.
Integration by parts will do. 
$$\int _0^1 x^m(1-x)^ndx  =$$
$$(1-x)^n \frac {x^{m+1}}{m+1}|_0^1 -\int _0^1\frac {x^{m+1}}{m+1}n(1-x)^{n-1}(-1)dx=$$
$$\frac{n}{m+1}\int _0^1 x^{m+1}(1-x)^{n-1}dx$$ 
Upon substitution you get the desired result. 
