I'd like to give another answer. Instead of the "repeated integration by parts" in the other answer, we can do the following:
We know the definition of the gamma function to be as follows:
$$\Gamma(s) = \int_{0}^\infty x^{s-1}e^{-x}dx$$
Now $\int_0^\infty e^{tx}\frac{1}{\Gamma(s)}\lambda^sx^{s-1} e^{-x\lambda}dx$ = $\frac{\lambda^s}{\Gamma(s)}\int_0^\infty e^{(t-\lambda)x}x^{s-1}dx$. We then integrate by substitution, using $u = (\lambda - t)x$, so also $x=\frac{u}{\lambda-t}$. This gives us $\frac{du}{dx}=\lambda - t$, i.e. $dx = \frac{du}{\lambda-t}$. Now let's put this into the integral, so then we get:
$$\frac{\lambda^s}{\Gamma(s)}\int_0^\infty e^{-u}{\left(\frac{u}{\lambda -t}\right)}^{s-1}\frac{du}{\lambda-t} = \left(\frac{\lambda}{\lambda-t}\right)^s\frac{1}{\Gamma(s)}\int_0^\infty u^{s-1}e^{-u}du$$
Here on the right-hand side we recognize the integral as the gamma function, so we get $\left(\frac{\lambda}{\lambda-t}\right)^s\frac{\Gamma(s)}{\Gamma(s)}$. Those gamma's cancel so then we get what we want:
$$\left(\frac{\lambda}{\lambda-t}\right)^s$$
Best Answer
Yes, you are right.
For that special type of gamma distribution, another name for it is Erlang distribution.
It is the formula of the sum of $n$ independent exponential distributions, hence we need to raise the power.