Don’t think you could derive a similar result because exponentiation is not linear like multiplication. Your result follows from

∑n^a =: S,

∑kn^a = k∑n^a = kS

and

kS = S ⇒ [S = 0] ∪ [k = 1].

But

∑(n^(a))^k = ∑n^(ka),

from where you could conclude that

k = 1 ⇒ ∑n^(ka) = ∑n^a = S

(the converse might not be true; there might be other cases which give

∑n^(ka) = S

without k equalling 1). Not sure the S = 0 bit would hold, and I can’t think of another result you could derive from there.