Prove that $I_{m, n} = I_{m+1,n-1}$ for $n\geq 1$.

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}