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
kS = S ⇒ [S = 0] ∪ [k = 1].
∑(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.