Let $X_1,\dots,X_{k+1}$ be mutually independent random variables, each having a gamma distribution with parameters $\alpha_i,i=1,2,\dots,k+1$ show that $Y_i=\frac{X_i}{X_1+\cdots+X_{k+1}},i=1,\dots,k$, have a joint ditribution as $\text{Dirichlet}(\alpha_1,\alpha_2,\dots,\alpha_k;\alpha_{k+1})$
According to the above theorem, we can get Dirichlet random variables from Gamma distribution.
So, I think that it's possible to calculate log likelihood of Dirichlet Distribution using Gamma.
Let $\vec{X} = (X_1,X_2,...,X_n)$ be samples from $Gamma(\alpha_i, 1)$. Then we normalize $\vec{X}$ and get $\vec{Y}$.
Then, Let $A = log\,p(\vec{X}) = \sum_{i=1}^nlog\,p(X_i) = \sum_{i=1}^n [(\alpha_i-1)\,log\,X_i - X_i - log\,\Gamma(\alpha_i)]\,$
$B = log\,p(\vec{Y})=\Sigma_{i=1}^n[(\alpha_i-1)log\,Y_i-log\,\Gamma(\alpha_i)]+log\,\Gamma(\bar{\alpha}) = \Sigma_{i=1}^n[(\alpha_i-1)(logX_i-log\bar{X})-log\,\Gamma(\alpha_i)]-log\,\Gamma(\bar{\alpha})$.
With the calculation, I think the result says that $A \neq B$. I don't know why this happen. I think that dirichlet random variable $\vec{Y}$ comes from Gamma random variable $\vec{X}$. So, I think they have the same log likelihood, but they didn't. Would you expalin about this?