How do I start this integral problem: $\int_0^1 \frac{\ln^3 u}{1-u} du = -\frac{\pi^4}{15} $? How do I prove this?
$$\int_0^1 \frac{\ln^3 u}{1-u} du = -\frac{\pi^4}{15} $$ 
I'm guessing using Riemann zeta function? But then how do I start? 

$$
\begin{align*}
\int_0^1 \frac{\ln^3(x)}{1-x}dx &= \int_0^1\ln^3(x)\sum_{n=0}^\infty x^n \; dx\\ 
&= \sum_{n=0}^\infty \int_0^1  x^n \ln^3(x) \; dx \\
&= -6\sum_{n=0}^\infty \frac{1}{(n+1)^4} \\
&= -6 \zeta(4) \\
&= -6 \times \frac{\pi^4}{90} \\ 
&= -\frac{\pi^4}{15}
\end{align*}
$$
