Is it reasonable to assert that $0^k - 1^k - 2^k + 3^k - 4^k + 5^k + 6^k - 7^k - ... = 0$ for all $k > 1$?

Here the signs are given by the Thue-Morse sequence; that is, the sign of $m^k$ is $+$ or $-$ according to whether the number of 1's in the binary expansion of $m$ is even or odd.

I'm aware that the theory of divergent series is actually an assortment of theories that don't always agree (even though they usually do), so that it isn't necessarily meaningful to ask what "the" value of a divergent series is without specifying what sort of regularization one has in mind. Still, the following reasoning strikes me as having some evidentiary force.

Take $f(q):=(1-q)(1-q^2)(1-q^4)(1-q^8)\cdots$ for $0 \leq q \leq 1$. Because of the numerous factors of $1-q$, we have $f^{(k)}(1) = 0$ for all $k \geq 0$. (Fussy technicality: $f^{(k)}(1)$ needs to be defined as a one-sided derivative because $f(q)$ is undefined for $q>1$.) If we repeatedly differentiate $f(q)$ term-by-term, and then plug in $q=1$, we get divergent sums of the form $p(0) - p(1) - p(2) + p(3) - p(4) + p(5) + p(6) - p(7) - ...$ where $p(\cdot)$ is a polynomial; since $f^{(k)}(1) = 0$, we are "justified" in asserting that the divergent sum vanishes. Since these polynomials form a basis for the space of all polynomials, we have $p(0) - p(1) - p(2) + p(3) - p(4) + p(5) + p(6) - p(7) - ... = 0$ for all polynomials $p(x)$, and hence in particular for $p(x) = x^k$.

Compare with the fairly well-known fact (source?) that if $n=2^d$ then $0^k - 1^k - 2^k + 3^k - ... \pm (n-1)^k$ vanishes for all $k$ between 0 and $d-1$.