How do you evaluate a sum of the form $\sum\limits_{n=0}^\infty nk^n$, for $|k|<1$?
If we take $k=\frac{1}{2}$, the sum is $\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \dots$, which isn't like any series I know. Wolfram Alpha evaluates this to $\frac{k}{(k-1)^2}$, but I don't know how it finds that.
The power series $$ \frac{1}{1 - z} = \sum_{n \geq 0} z^n$$ converges absolutely for all $|z| < 1$. Some results about power series also say that it will be well-behaved under differentiation in that region, so differentiate both sides. $$ \frac{1}{(1-z)^2} = \sum_{n \geq 0} n z^{n-1}$$ which is almost, but not quite, the series you want. Multiplying both sides by $z$ fixes this: $$ \frac{z}{(1 - z)^2} = \sum_{n \geq 0} n z^n$$ which is the closed form that Wolfram Alpha gave.
Consider $S(n)=\sum_{n=0}^{\inf}k^n$. If you take the derivative of $S(n)$ (we are allowed to do this because this is a power series with $|k|<1$), you will notice that you get another infinite sum with the slight modification: $$K(n)=\sum_{n=0}^{\inf}(n+1)k^n$$. Notice that the sum of your interest, $$L(n)=k.K(n).$$ Since $S(n)$ is an infinite geometric series that converges, you can calculate this explicitly and that will lead you to your answer.
Or you can choose to use induction since you know the answer.
If you don't want to use derivative of power series: \begin{align} \frac{z-z^{n+1}}{1 - z}& =& &z+&z^2+&z^3+...+&z^n \\ \frac{z^2-z^{n+1}}{1 - z}& =& &&z^2+&z^3+...+&z^n \\ \frac{z^3-z^{n+1}}{1 - z}& =& &&&z^3+...+&z^n \\ \frac{z^n-z^{n+1}}{1 - z}& =& &&&&z^n \\ \end{align}
Adding them together you get $$\sum_{k=1}^n kz^k = \frac{z+z^2+..+z^n-nz^{n+1}}{1-z}= \frac{z\frac{1-z^{n}}{1-z}-nz^{n+1}}{1-z}=\frac{z-(n+1)z^{n+1}+nz^{n+2}}{(1-z)^2}$$
Now, use the fact tha for $|z|<1$ we have $$\lim_n (n+1)z^{n+1}=\lim_n nz^{n+2}=0$$
