Distribution function Find (without using MGF) the mean and variance.
$$f(x) = \exp(-kx)x^{(r-1)}k^r/(r-1)!\  \text{ for }\ x>=0$$
$$f(x) = 0\ \text{ for }\ x<0$$
$r$ positive integer, $k>0$
 A: The answer is actually already there in the question, staring us in the face.  All we need to do is interpret it appropriately.
Look at the constants in the definition of $f$: $k^r$ divided by $(r-1)!$. The first factor should really be absorbed by the $x$'s and the implicit $dx$.  That is,
$$k^r x^{r-1} dx = (kx)^{r-1} d(kx).$$
This makes it clear that the distribution is really a function of $kx$, revealing $k$ as a scale factor.  If we can find the mean $\mu$ and variance $\sigma^2$ for $k=1$, we would only need to divide them by $k$ and $k^2$, respectively, to get the fully general answer.
That observation reduces the question to one of finding various moments of 
$$f_r(x) = \frac{x^r \exp(-x)}{x (r-1)!}.$$
There's only one constant in there, and it must be there to make sure this distribution integrates to unity.  (That is the import of @jbowman's hint in the comments.)  In other words, you already know that
$$\int_0^\infty x^r \exp(-x) \frac{dx}{x} = (r-1)!.$$
To find the $j$th moment ($j=1,2$ are all that are needed), you must obtain a formula for
$$\int_0^\infty x^j f_r(x) dx = \int_0^\infty  x^j\left(x^r \exp(-x) \frac{dx}{x}\right) = \int_0^\infty x^{r+j} \exp(-x) \frac{dx}{x}.$$
It is now but a moment's work to apply the penultimate formula to this expression to obtain the answers.  I leave the details--which are simple algebra, since the integration has been done--to the interested reader.
