Algebra. Solving for a given gamma function: $\ln L = n \ln(\Gamma(a+1)) - n \ln (\Gamma(a)) + (a-1) \sum_{i=1}^{n} \ln x_i$ $\ln L = n \ln(\Gamma(a+1)) - n \ln (\Gamma(a)) + (a-1) \sum_{i=1}^{n} \ln x_i$
so given this I want to solve the derivative for $a$ then solve for $a$, $\ln L = 0$
$0 = \frac{n(\Gamma(a+1)')}{\Gamma(a+1)} - \frac{n\Gamma'(a)}{\Gamma(a)} + \sum_{i=1}^{n} \ln x_i$
it equals to $\frac{n}{a} + \sum_{i=1}^{n} \ln x_i$ then $a = -\frac{n}{\sum_{i=1}^{n} \ln x_i}$. Not sure how they got to that though.
Not sure how to do the rest.

It's simply that
$$ \frac{\Gamma(a + 1)}{\Gamma(a)} = a,$$
by a standard property of the Gamma function.
So
$$ \ln L = n \ln a + (a - 1) \sum_{i=1}^n \ln x_i.$$
