Intruiging Symmetric harmonic sum $\sum_{n\geq 1} \frac{H^{(k)}_n}{n^k}\, = \frac{\zeta{(2k)}+\zeta^{2}(k)}{2}$

Intruiging Symmetric harmonic sum $\sum_{n\geq 1} \frac{H^{(k)}_n}{n^k}\,
= \frac{\zeta{(2k)}+\zeta^{2}(k)}{2}$

I proved the following equation
$$\sum_{n\geq 1} \frac{H^{(k)}_n}{n^k}\, =
\frac{\zeta{(2k)}+\zeta^{2}(k)}{2}$$
We define
$$H^{(k)}_n=\sum_{m= 1}^n \frac{1}{m^k}$$
I am looking forward to seeing what approaches would you use .