Integrating by parts repeatedly to find factorial recurrence I'm trying to solve this problem, but I'm stuck with the recursion, after repeatedly integrating by parts I found elements of the factorials of the answer. However, I don't know how to factorize it to avoid cancelling everything when evaluating the bounds.

Let $ 1 \le i < j \le n$ and integrating by parts repeatedly prove
that:
$$\int_0^z y^i (z-y)^{j-i-1} \, dy = \frac{i! (j-i-1)!}{j!}z^j$$
$$\int_0^1 (1-z)^{n-j} z^{j+1} \, dz = \frac{(j+1)!)(n-j)!}{(n+2)!} $$

 A: Let $I(i,j)=\int_0^z y^i(z-y)^{j-i-1}dy$. Then using integration by part,
\begin{align*}
I(i,j)&=\int_0^z y^i(z-y)^{j-i-1}dy\\
&=\frac{-1}{j-i}\int_0^z y^i d(z-y)^{j-i}\\
&=\frac{-1}{j-i}\left[y^i(z-y)^{j-i}\right]_0^z+\frac{1}{j-i}\int_0^z (z-y)^{j-i}\cdot iy^{i-1}dy\\
&=\frac{i}{j-i}\int_0^z y^{i-1}(z-y)^{j-i}dy\\
&=\frac{i}{j-i}I(i-1,j+1)
\end{align*}
Now an inductive argument yields
\begin{align*}
I(i,j)&=\frac{i}{j-i}I(i-1,j+1)\\
&=\frac{i}{j-i}\cdot\frac{i-1}{j-i+1}I(i-2,j+2) &&(*)\\
&=\frac{i}{j-i}\cdot\frac{i-1}{j-i+1}\cdot\frac{i-2}{j-i+2}I(i-3,j+3)\\
&=\cdots\\
&=\frac{i}{j-i}\cdot \frac{i-1}{j-i+1}\cdot \frac{i-2}{j-i+1}\cdot\cdots\cdot \frac{1}{j-i+(i-1)}I(0,j+i)\\
&=\frac{i!}{(j-i)(j-i+1)(j-i+2)\cdots(j-i+(i-1))}\int_0^z (z-y)^{j-1}dy\\
&=\frac{i!}{(j-i)(j-i+1)(j-i+2)\cdots(j-1)}\cdot\frac{-1}{j}\left[(z-y)^j\right]_0^z\\
&=\frac{i!}{(j-i)(j-i+1)(j-i+2)\cdots(j-1)\cdot j}z^j\\
&=\frac{i!(j-i-1)!}{j!}z^j
\end{align*}
The second equality can be obtained by some suitable substitutions from the first equality, given how similar they are
$(*)$: Here is a WRONG inductive step one might get into:
$$I(i,j)=\frac{i}{j-i}I(i-1,j+1)=\frac{i}{j-i}\cdot\frac{i-1}{(j+1)-(i-1)}I(i-2,j+2)$$
This is actually INCORRECT because the denominator comes from integration by part, which has nothing to do with the exponent of $y$ (it comes from the exponent of $(z-y)$). If you are still doubt about this, try to continue the first part of the proof to find a recurrence between $I(i,j)$ and $I(i-2,j+2)$
