Prove that $\int_0^1\frac{\ln(1-x)\ln^2x}{x-1}dx=\frac{\pi^4}{180}$ Prove that (please)

$$\int_0^1\frac{\ln(1-x)\ln^2x}{x-1}dx=\frac{\pi^4}{180}$$

I've tried using Taylor series and I ended up with
$$-\sum_{m=0}^\infty\sum_{n=1}^\infty\frac{2}{n(m+n+1)^3}$$
I am stuck there and I couldn't continue it using partial fraction to get any familiar sum of series. Could anyone here please help me to prove the integral preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.

The idea is to use the fact that $\dfrac{\ln(1-x)}{x-1} = \sum\limits_{n \geq 1} H_n x^n$ where $|x| < 1$ and $H_n$ is the $n$'th harmonic number. This reduces the integral to a linear combination of $\sum\limits_{n \geq 1} H_n/n^3$ and $\sum\limits_{n \geq 1} 1/n^4$. The latter term is $\zeta(4) = \pi^4/90$; the former term is more difficult to calculate, but can be calculated with residues as well; a proof can be found here here. 
It seems very difficult to do this using only elementary methods.
More explicitly:
$$
I = \int_0^1\frac {\log(1 - u)}{u - 1}\, (\log u)^2 \, du = \int_0^1 \sum_{n = 0}^\infty H_n u^n \, (\log u)^2 \, du
= \sum_{n = 0}^\infty H_n \int_0^1 u^n\, (\log u)^2\, du = \sum_{n = 0}^\infty H_n\, \frac {2}{(n + 1)^3} = 2\sum_{n = 1}^\infty \frac {H_{n - 1}}{n^3}
= 2\left(\sum_{n = 1}^\infty \frac {H_n}{n^3} - \sum_{n = 1}^\infty \frac {1}{n^4}\right) = 2\left(\frac54\zeta(4) - \zeta(4)\right) = \frac{\pi^4}{180}.$$
Here we used the fact that $\int_0^{1}dx\, x^n \ln^2 x = \int_0^{\infty} dx\, e^{-(n+1)x} x^2 = \frac{2!}{(n+1)^3}$.
The penultimate equality is not so trivial to show (but see the comment by Tunk-Fey), but a proof is given in the link.
