Ways to put numbered balls in boxes no box being empty

You can do this by inclusion/exclusion. There are $\binom r0r^n$ ways of putting the $n$ balls into the $r$ boxes. From this we have to subtract the $\binom r1(r-1)^n$ ways of putting the $n$ balls into just $r-1$ of the boxes. To this we have to add the $\binom r2(r-2)^n$ ways of putting the $n$ balls into just $r-2$ of the boxes, and so on, so

$$a_n=\sum_{k=0}^r\binom rk(-1)^{r-k}k^n=\left.\left(q\frac{\partial}{\partial q}\right)^n(q-1)^r\right|_{q=1}\;.$$

This is one of the problems (counting surjective functions from N to X) in the so-called Twelvefold way (the other 11 problems might interest you as well). The solution is $r!\left\{n\atop r\right\}$, where $\left\{n\atop r\right\}$ denotes the Stirling number of the second kind written $S(n,r)$ in the answer by leonbloy. If you need to compute these numbers explicitly, it is more efficient to use the recurrence $\left\{n+1\atop r\right\} = r \left\{n \atop r\right\} + \left\{n\atop r-1\right\}$ for $1<r<n$ with boundary conditions $\left\{n\atop1\right\} = \left\{n\atop n\right\} = 1$, than to use the summation formulas given in other answers (but don't forget to multiply by $r!$ afterwards).