The probability density function for the gamma distribution is
$$f(x;a,b) = \frac{1}{b^a\Gamma(a)} x^{a-1}e^{-x/b},$$
so the integral we must consider is
$$\mathbb{E}(\ln (X^n)) = \frac{1}{b^a \Gamma(a)}
\int_0^\infty dx\, x^{a-1} e^{-x/b} \ln x^n.$$
Let $z=x/b$.
We find
$$\begin{eqnarray*}
\mathbb{E}(\ln (X^n))
&=& \frac{n}{\Gamma(a)} \int_0^\infty dz\, z^{a-1} e^{-z}\ln (b z) \\
&=& n\ln b + \frac{n}{\Gamma(a)} \int_0^\infty dz\, z^{a-1} e^{-z}\ln z \\
&=& n\ln b + \frac{n}{\Gamma(a)}
\frac{d}{d a} \int_0^\infty dz\, z^{a-1} e^{-z} \\
&=& n\left(\ln b + \frac{\Gamma'(a)}{\Gamma(a)}\right) \\
&=& n\left(\ln b + \psi(a)\right),
\end{eqnarray*}$$
where we have used the definition of the gamma function,
$\Gamma(a) = \int_0^\infty d x \, x^{a-1} e^{-x}$,
and the digamma function, $\psi(z) = \Gamma'(z)/\Gamma(z)$.
Addendum 1: It is possible to show (using the fact that $\ln x\leq z^s/s$ for all $s>0$) that
$\int_0^\infty dz\, z^{a-1} e^{-z}\ln z$
converges uniformly for $1< a < \infty$.
This justifies the trick of differentiating $\Gamma(a)$.
Addendum 2:
There is a possibility that the OP is interested in $\mathbb{E}((\ln X)^n)$.
@Sasha dealt with this below.
It can also be handled with the derivative trick,
$$\begin{eqnarray*}
\mathbb{E}((\ln X)^n)
&=& \frac{1}{\Gamma(a)} \int_0^\infty dz\, z^{a-1} e^{-z}(\ln (b z))^n \\
&=& \frac{1}{\Gamma(a)} \int_0^\infty dz\, z^{a-1} e^{-z}(\ln b + \ln z)^n \\
&=& \frac{1}{\Gamma(a)} \int_0^\infty dz\, z^{a-1} e^{-z}
\sum_{k=0}^n {n \choose k} (\ln b)^{n-k} (\ln z)^k \\
&=& \frac{1}{\Gamma(a)} \sum_{k=0}^n {n \choose k} (\ln b)^{n-k}
\int_0^\infty dz\, z^{a-1} e^{-z} (\ln z)^k \\
&=& \frac{1}{\Gamma(a)} \sum_{k=0}^n {n \choose k} (\ln b)^{n-k}
\frac{d^k}{d a^k} \int_0^\infty dz\, z^{a-1} e^{-z} \\
&=& \frac{1}{\Gamma(a)} \sum_{k=0}^n {n \choose k} (\ln b)^{n-k}
\Gamma^{(k)}(a).
\end{eqnarray*}$$
This is equivalent to the solution given by @Sasha.
For $n=2$ we arrive at
$$\mathbb{E}((\ln X)^2) = (\ln b + \psi(x))^2 + \psi^{(1)}(a),$$
where $\psi^{(m)}(z) = d^{m+1}(\ln \Gamma(z))/d z^{m+1}$ is the polygamma function.