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!

*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.
