Show that $\int\cdots\int{p_1^{x_1}\cdots p_m^{x_m} \, dp_1\cdots dp_m}=\frac{x_{1}!\cdots x_{m}!}{\left(m-1+\sum_{i=1}^{m}{x_i}\right)!}$ 
How to prove that,$$\int\int\cdots\int p_1^{x_1} p_2^{x_2} \cdots p_m^{x_m} \, dp_1 \, dp_2 \cdots dp_m = \frac{x_1!\cdots x_m!}{\left(m-1+\sum_{i=1}^m x_i\right)!}$$
  where $0\leq p_i\leq 1$ and $\sum\limits_{i=1}^m p_i=1$, 

My attempt: If we have, $$\int_{0}^{1}\int_{0}^{1}{p_{1}^{x_{1}}p_{2}^{x_{2}}dp_{1}dp_{2}}=\int_{0}^{1}\int_{0}^{1}{p_{1}^{x_{1}}(1-p_{1})^{x_{2}}dp_{1}dp_{2}}$$
$$=\int_{0}^{1}{p_{1}^{x_{1}}(1-p_{1})^{x_{2}}dp_{1}}=\beta(x_{1}+1,x_{2}+1)=\dfrac{x_{1}!x_{2}!}{(x_{1}+x_{2}+2)!}$$
Since, $p_{1}+p_{2}=1$ and $\beta(x,y)=\dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$. Now, for the next case
$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}{p_{1}^{x_{1}}p_{2}^{x_{2}}p_{3}^{x_{3}}dp_{1}dp_{2}dp_{3}}=\int_{0}^{1}\int_{0}^{1}{p_{1}^{x_{1}}p_{2}^{x_{2}}(1-(p_{1}+p_{2}))^{x_{3}}dp_{1}dp_{2}}$$
I tried with change variable $p_{1}+p_{2}=t$, but doesn't work.  Any help, for prove the general statement... Regards!
 A: The case $m=2$ is just the definition of the beta function:
$$ \int_{0}^{1} x^a(1-x)^b\,dx = \frac{\Gamma(a+1)\cdot\Gamma(b+1)}{\Gamma(a+b+2)} = \frac{a!\cdot b!}{(a+b+1)!} $$
while the general case is the normalizing constant of the Dirichlet distribution.
In order to prove the general statement you just have to follow the lines outlined in the paragraph Relationship between gamma function and beta function:


*

*Write $x_j!=\Gamma(x_j+1)$ as $\int_{0}^{+\infty}u_j^{x_j} e^{-u_j}\,du_j$ and multiply that integrals in order to get an integral over $(\mathbb{R}^+)^m$;

*Apply a change of variable $u_1+\ldots+u_m=t$;

*Apply the Fubini theorem in order to get the original integral multiplied by some value of the $\Gamma$ function, namely $\Gamma(m-1+x_1+\ldots+x_m)$. 

A: *I know this question has existed for a long time. It happens that I also came across this formula learning Bose-Einstein distribution and was confused for a moment. However, it turns out that the step-by-step calculation wasn't that daunting for some lower dimension case. And indeed the whole formula can be derived by using induction.
Let us embark on calculating $\int_{0}^{1}\int_{0}^{1} p_1^{x_1}p_2^{x_2}(1-p_1-p_2)^{x_3} dp_1dp_2 $.
Start with a change of variable $p_1+p_2=t$, we have $\int_{0}^{1}\int_{0}^{t} p_1^{x_1}(t-p_1)^{x_2}(1-t)^{x_3} dp_1dt$
= $\int_{0}^{1}(1-t)^{x_3} (\int_{0}^{t} p_1^{x_1}(t-p_1)^{x_2}dp_1)dt$ (1.1)
Now, $\int_{0}^{t} p_1^{x_1}(t-p_1)^{x_2}dp_1$ do look like the $\int_{0}^{1} p_1^{x_1}(1-p_1)^{x_2}dp_1$, and if you remember the last formula, you know that
$\int_{0}^{1} p_1^{x_1}(1-p_1)^{x_2}dp_1 = \frac{x_1!x_2!}{(x_1+x_2+1)!}$ either by the Beta function or recursively using integration by parts.
Note that we have used the condition that $x_1+x_2 = n$.
You have very good reason to guess $\int_{0}^{t} p_1^{x_1}(t-p_1)^{x_2}dp_1$ can be calculated in an identical manner. Consider the base case, where we have
$\int_{0}^{t} (t-p_1)^{m-x_3}dp_1 = \int_{0}^{t} (t-p_1)^{x_1+x_2}dp_1 = \frac{1}{x_1+x_2+1}t^{x_1+x_2+1}$
Now $\int_{0}^{t} p_1^{1}(t-p_1)^{x_1+x_2-1}dp_1 = -\frac{1}{x_1+x_2}\int_{0}^{t} p_1d(t-p_1)^{x_1+x_2} = -\frac{1}{x_1+x_2}[p_1(t-p_1)^{x_1+x_2}|_{0}^{t}-\int_{0}^{t} (t-p_1)^{x_1+x_2}dp_1]$
= $\frac{1}{(x_1+x_2)(x_1+x_2+1)}t^{x_1+x_2+1}$, you can continue to do for another step so that the pattern is clear, but here we have
$\int_{0}^{t} p_1^{x_1}(t-p_1)^{x_2}dp_1 = \frac{x_1!x_2!}{(x_1+x_2+1)!}t^{x_1+x_2+1}$, plug this back to (1.1), and you will find the desired result.
Use this idea, you can prove by induction your statement in a general case.
