A series involving the harmonic numbers : $\sum_{n=1}^{\infty}\frac{H_n}{n^3}$

Let $H_{n}$ be the nth harmonic number defined by $ H_{n} := \sum_{k=1}^{n} \frac{1}{k}$.

How would you prove that

$$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac{\pi^4}{72}?$$

Simply replacing $H_{n}$ with $\sum_{k=1}^{n} \frac{1}{k}$ does not seem like a good starting point. Perhaps another representation of the nth harmonic number would be more useful.