I personally prefer a derivation using the principle of virtual work where the formula of torque directly comes out. While angular momentum is a natural property to consider for a spherically symmetrical problem, this alternative approach shows its relevance for statics of rigid bodies even when this symmetry is not present.  

Take a set of points indexed by $i$ at position $\vec r_i$, on which are applied respectively the forces $\vec F_i$. This gives first formula of vitual work for a general displacement:
$$
\delta W = \sum \vec F_i \cdot \delta\vec r_i
$$
Furthermore, lets assume the points are rigidly constrained and can only rotate around the origin. Any allowed differential displacement can thus be written as $\delta\vec r_i =\delta\vec \phi \times \vec r_i$ where $\vec \phi$ is the differential angular displacement. Injecting in the work you get:
$$
\begin{align}
\delta W &= \sum \vec F_i \cdot (\delta\vec \phi \times \vec r_i)\\
&= (\sum \vec r_i \times\vec F_i ) \cdot \delta\vec \phi
\end{align}
$$

So static equilibrium is equivalent to a vanishing virtual work for any relevant virtual displacement, hence $\sum \vec r_i \times\vec F_i $, the torque naturally pops out. It also explains also the useful power formula for rotation (with angular velocity $\vec \omega$):

$$
P = \left(\sum \vec r_i \times\vec F_i \right) \cdot \vec \omega
$$

Hope this helps and tell me if you find some mistakes.