It is possible to evaluate it by converting it into a double summation and reversing the order of summation:
$$\begin{align*}
\sum_{k=0}^nka^k&=\sum_{k=1}^nka^k\\
&=\sum_{k=1}^n\sum_{j=1}^ka^k\tag{1}\\
&=\sum_{1\le j\le k\le n}a^k\tag{2}\\
&=\sum_{j=1}^n\sum_{k=j}^na^k\tag{3}\\
&=\sum_{j=1}^n\frac{a^j-a^{n+1}}{1-a}\\
&=\sum_{j=1}^n\frac{a^j}{1-a}-\sum_{j=1}^n\frac{a^{n+1}}{1-a}\\
&=\frac1{1-a}\sum_{j=1}^na^j-\frac{na^{n+1}}{1-a}\\
&=\frac1{1-a}\cdot\frac{a-a^{n+1}}{1-a}-\frac{na^{n+1}}{1-a}\\
&=\frac{a-a^{n+1}}{(1-a)^2}-\frac{na^{n+1}}{1-a}
\end{align*}$$
The summations $(1),(2)$, and $(3)$ are just three different ways of writing a sum over all pairs of indices $j$ and $k$ satisfying the inequalities $1\le j\le k\le n$.
If you examine this closely, you’ll see that it’s really just what bof did in his answer: he split $ka^k$ into $k$ copies of $a^k$ arranged in a column, and then instead of summing each column to get the sums $a^1,2a^2,\ldots,na^n$ and then adding those sums, he summed the rows, which were geometric series and hence easy to sum, and then added these row sums. In other words, he reversed the order of summation, just as I did. His is the presentation that I’d use to introduce the idea; mine shows how it can be expressed more compactly and how you might see it written if some familiarity with summation notation is assumed (though I did include a few extra steps than usual for greater clarity).