It is possible to get a closed-form for $1+2^i+3^i+\cdots (N-1)^i$? Let $i=\sqrt{-1}$ the complex imaginary unit, taking $$arg(2)=0$$ for the definition of the summand $2^i$ in $$1^i+2^i+3^i+\cdots (N-1)^i,$$
as $$2^i=\cos\log 2+ i\sin\log 2,$$
see [1].

Question. It is possible to get a closed-form (or the best approximation possible), for an integer $N\geq 1$ 
  $$1+2^i+3^i+\cdots (N-1)^i,$$
  where the summands are defined in the same way, taking principal branches of complex argument and complex exponentiation? 

Thanks in advance, my goal is start to refresh some easy facts in complex variable, please tell me if there are mistakes in the use of previous  definitions.
References:
[1] MathWorld, http://mathworld.wolfram.com/ComplexExponentiation.html http://mathworld.wolfram.com/ComplexArgument.html
 A: I think there is a closed form, notice:
$$1^i+2^i+3^i+\dots+(n-1)^i=\sum_{k=2}^{n}\left(k-1\right)^i=\sum_{k=1}^{n-1}k^i=\text{H}_{n-1}^{(-i)}=\zeta(-i)-\zeta(-i,n)$$
Where $\zeta(s,a)$ is the Hurwitz zeta function, $\zeta(s)$ is the Riemann zeta function and 
$\text{H}_{n}^{(r)}$ is the generalized harmoic number.

EDIT:
$$\zeta(-i)=\lim_{s\to0}\left[\sum_{k=1}^{\infty}k^{i-s}\right]\approx 0.0033+0.4182i$$
A: First of all, you make the assumption $$1^i=1$$When this is not true.$$1^i=e^{\pm2\pi n},n=0,1,2,3,\ldots$$
More generally, I will solve$$1^x+2^x+3^x+4^x+5^x\ldots=\sum_{n=1}^{\infty}n^x$$This has solution obtainable via permutation:$$\sum_{n=2}^{m}n^x+n=\sum_{n=1}^{m-1}n^x+n^m$$$$\sum_{n=1}^{m-1}(n+1)^x+n=\sum_{n=1}^{m-1}n^x+n^m$$Apply Binomial thereom:$$(n+1)^x=\sum_{n=0}^{\infty}\frac{n^{x-n}1^nx!}{(x-n)!n!}$$$$\sum_{n=1}^{m-1}\sum_{i=n}^{\infty}\frac{n^{x-n}1^nx!}{(x-n)!n!}+n=\sum_{n=1}^{m-1}n^x+n^m$$
And now, to be honest with you, I can't proceed from here.
Or perhaps I took it from the wrong method of solving.
