Mathematical justification for Le Chatelier's principle
I think that I might have an answer to my own question:

$$K=exp \left( \frac{TΔS^o-ΔH^o}{RT} \right) $$

implies that K is dependent on T, but not on total pressure of system, p (nor concentrations).

For
$$ A_{(g)}⇌2B_{(g)} $$
$$ K= \frac{{p_{B}}^2}{p_{A}} $$
Partial pressure of a gas = mole fraction (mf) x total pressure of system

If p is increased then K initially increases:
$$ K=\frac{(mf_B.p)^2}{mf_A.p} $$
$$ ⇒ K=\frac{{mf_B}^2.p}{mf_A} $$

But p doesn't affect K, yet K has increased. So the mole fraction of A must be increased by shifting the position of equilibrium to the left (thus decreasing the mole fraction of B), in order to re-establish the value of K to before p was increased.