In the proof for the Dirichlet Distribution, I found the following step:
$$f_{M-1}=\int^{\mu_{M-1}}_{0}\bigg(\mu_{M-1}^{(\alpha_{m-1})-1}\bigg)\bigg(1-\sum_{j=1}^{M-1}\mu_j\bigg)d\mu_{M-1}~~~~~~~~(1)$$
In order to make the limits of integration equal to 0 and 1, we substitute:
$$\mu_{M-1}=t\bigg(1-\sum_{j=1}^{M-2}\mu_j\bigg)$$
Somehow, this results in the integral:
$$f_{M-1}=\bigg(1-\sum_{j=1}^{M-2}\mu_j \bigg)^{(\alpha_{M-1})+(\alpha_M)-1}\int_{0}^{1} t^{(\alpha_{M-1})-1}(1-t)^{{\alpha_M}-1}dt~~~~~~~~(2)$$
$$f_{M-1}=\bigg(1-\sum_{j=1}^{M-2}\mu_j \bigg)^{(\alpha_{M-1})+(\alpha_M)-1}\frac{\Gamma(\alpha_{M-1})\Gamma(\alpha_{M})}{\Gamma(\alpha_{M-1}+\alpha_{M})}~~~~~~~~(3)$$
What's the technique to get from equation (1) to equation (2)?
Also, we know that:
$$\sum^{M}_{k=1} \mu_k = 1$$
Big hint:
$$\bigg(1-\sum_{j=1}^{M-1} \mu_j\bigg)=\bigg(1-t\bigg)\bigg(1-\sum_{j=1}^{M-2}\mu_j\bigg)$$