# How to normalize Dirichlet distribution?

The Dirichlet distribution is defined as: $$p(\vec{\mu}_M|\vec{\alpha}_M) = c(\vec{\alpha}_M) \Pi_{k=1}^M\mu_k^{\alpha_k-1}$$ where $$\vec{\mu}_M, \vec{\alpha}_M$$ is a vector of length $$M$$ and $$\sum_{k=1}^M\mu_k=1$$.

I want to show that: $$c(\vec{\alpha}_M)=\frac{\Gamma(\alpha_1+\alpha_2+\cdots+\alpha_M)}{\Gamma(\alpha_1)\Gamma(\alpha_2)\cdots\Gamma(\alpha_M)}$$

Prove by induction: when $$M=2$$, the distribution is the same as Beta distribution, the relation holds.

Suppose it holds $$M=N-1$$, when $$M=N$$:

\begin{align} 1=&\int p(\vec{\mu}_N|\vec{\alpha}_N) \\ =& \vec{c}(\vec{\alpha}_N) \int \mathrm{d}\vec{\mu}_N \Pi_{k=1}^N\mu_k^{\alpha_k-1} \\ =& \vec{c}(\vec{\alpha}_N) \int_{0}^{1} \mathrm{d}\mu_N \mu_N^{\alpha_N-1}\int \mathrm{d}\vec{\mu}_{N-1}\Pi_{k=1}^{N-1}\mu_k^{\alpha_k-1} \end{align}

For $$\sum_{k=1}^{N-1}\mu_k = 1-\mu_N$$, if we change the variable $$u_k=\mu_k/(1-\mu_N)$$, then $$\sum_{k=1}^{N-1}u_k = 1$$.

Consider: \begin{align} &\int \mathrm{d}\vec{\mu}_{N-1}\Pi_{k=1}^{N-1}\mu_k^{\alpha_k-1} \\ =&(1-\mu_N)^{\sum_{k=1}^{k=N-1}\alpha_k} \int \mathrm{d}\vec{u}_{N-1}\Pi_{k=1}^{N-1}u_k^{\alpha_k-1} \\ =&(1-\mu_N)^{\sum_{k=1}^{k=N-1}\alpha_k} \frac{1}{c(\vec{\alpha}_{N-1})} \end{align} where I have used the assumption holds when $$M=N-1$$.

Therefore: \begin{align} 1=\frac{c(\vec{\alpha}_{N})}{c(\vec{\alpha}_{N-1})} \int_{0}^{1} \mathrm{d}\mu_N \mu_N^{\alpha_N-1} (1-\mu_N)^{\sum_{k=1}^{k=N-1}\alpha_k} \end{align}

It seems that I am very close to the desired result but missed a factor of $$1/(1-\mu_N)$$ in the final integrand, how to fix this?