Cancelling items in product over two indices $$ \int_0^1 x^n (1-x)^m dx$$
Do parts:
$$ I_{n.m} = \frac{1}{n+1}\left( \left[x^{n+1} (1-x)^m \right]_0^1 + m \int_0^1 x^{n+1} (1-x)^{m-1}\right)$$
Or,
$$ I_{n,m}= \frac{m}{n+1} \left(I_{n+1,m-1} \right)$$
Product both sides over m from $1 \to k$ (refer here)
$$ \prod_{m=1}^k I_{n,m} = \prod_{m=1}^k \frac{m}{n+1} I_{n+1,m-1}$$
Or,
$$ \prod_{m=1}^k I_{n,m} = m! \prod_{m=1}^k \frac{1}{n+1} I_{n+1,m-1} \tag{1}$$
Product both sides over $n$ from $1 \to j$,
$$ \prod_{n=1}^j\prod_{m=1}^k I_{n,m} = \prod_{n=1}^jk! \prod_{m=1}^k \frac{1}{n+1} I_{n+1,m-1} \tag{1}$$
Or,
$$ \prod_{n,m=1}^{j,k} I_{n,m} = \frac{(k!)^j}{(j!)^m} \prod_{n,m=1}^{j,k} I_{n+1,m-1}$$
The left is summed over the domain $n \times m= (1,j) \times (1,k)$ and the right is over the domain $n \times m = (2,j+1) \times (0,k-1)$. I need advice on how to cancel like terms, the cancellation part is really tripping me over.
 A: We consider $I_{n,m}=\int_{0}^1x^n(1-x)^m\,dx$ and the recurrence relation:
\begin{align*}
I_{n,m}&=\frac{m}{n+1}I_{n+1,m-1}\qquad\qquad n,m\geq 1\\
I_{n,0}&=I_{0,n}=\frac{1}{n+1}\qquad\qquad\quad n\geq 0\tag{1}
\end{align*}
Let $j,k$ be positive integers. We obtain
\begin{align*}
\prod_{n=1}^j\prod_{m=1}^k I_{n,m}&=\prod_{n=1}^j\prod_{m=1}^k \left(\frac{m}{n+1}I_{n+1,m-1}\right)\\
&=\prod_{n=1}^j\frac{k!}{(n+1)^k}\prod_{m=1}^k I_{n+1,m-1}\\
&=\frac{\left(k!\right)^j}{\left((j+1)!\right)^k}\prod_{n=1}^j\prod_{m=1}^k I_{n+1,m-1}\\
&=\frac{\left(k!\right)^j}{\left((j+1)!\right)^k}\prod_{n=2}^{j+1}\prod_{m=0}^{k-1} I_{n,m}\tag{2}\\
&=\frac{\left(k!\right)^j}{\left((j+1)!\right)^k}\left(\prod_{m=0}^{k-1}I_{j+1,m}\right)\prod_{n=2}^{j}\prod_{m=0}^{k-1} I_{n,m}\tag{3}\\
&=\frac{\left(k!\right)^j}{\left((j+1)!\right)^k}\left(\prod_{m=0}^{k-1}I_{j+1,m}\right)
\left(\prod_{n=2}^j I_{n,0}\right)\prod_{n=2}^{j}\prod_{m=1}^{k-1} I_{n,m}\tag{4}\\
&=\frac{\left(k!\right)^j}{\left((j+1)!\right)^k}\left(\prod_{m=0}^{k-1}I_{j+1,m}\right)
\left(\prod_{n=2}^j \frac{1}{n+1}\right)\prod_{n=2}^{j}\prod_{m=1}^{k-1} I_{n,m}\tag{5}\\
&\,\,\color{blue}{=\frac{2\left(k!\right)^j}{\left((j+1)!\right)^{k+1}}\left(\prod_{m=0}^{k-1}I_{j+1,m}\right)
\prod_{n=2}^{j}\prod_{m=1}^{k-1} I_{n,m}}\tag{6}\\
\end{align*}
Comment:

*

*In (2) we shift the indices $n$ and $m$ to derive the representation $I_{n,m}$ as preparation for cancellation.


*In (3) we separate the factors $I_{j+1,.}$.


*In (4) we separate the factors $I_{.,0}$.


*In (5) we use the identity (1).


*In (6) we use the identity $\prod_{n=2}^j\frac{1}{n+1}=\frac{2}{(j+1)!}$.
On the other hand we have
\begin{align*}
\color{blue}{\prod_{n=1}^j\prod_{m=1}^k I_{n,m}}&=\left(\prod_{n=1}^j I_{n,k}\right)\prod_{n=1}^j\prod_{m=1}^{k-1} I_{n,m}\tag{7}\\
&=\left(\prod_{n=1}^j I_{n,k}\right)\left(\prod_{m=1}^{k-1} I_{1,m}\right)\prod_{n=2}^j\prod_{m=1}^{k-1} I_{n,m}\tag{8}\\
&\,\,\color{blue}{=\left(\prod_{n=1}^j I_{n,k}\right)\left(\prod_{m=1}^{k-1}\frac{1}{m^2+3m+2}\right)\prod_{n=2}^j\prod_{m=1}^{k-1} I_{n,m}}\tag{9}
\end{align*}
Comment:

*

*In (7) we separate factors $I_{n,k}$ to get the upper index $j$ in the product as in (6).


*In (8) we separate factors $I_{1,m}$ to start with $n=2$ as in (6).


*In (9) we use the identity $I_{1,m}=\frac{1}{m^2+3m+2}$ valid for $m\geq 1$.

We can now cancel the double product in (6) and (9) and obtain
\begin{align*}
\color{blue}{\frac{2\left(k!\right)^j}{\left((j+1)!\right)^{k+1}}\prod_{m=0}^{k-1}I_{j+1,m}
=\prod_{m=1}^{k-1}\frac{1}{m^2+3m+2}\prod_{n=1}^j I_{n,k}}
\end{align*}

