The remainder when $1^1+3^3+5^5+\dots + 1023^{1023} \pmod{1024}$ is $\dots$ The remainder when $1^1+3^3+5^5+\dots + 1023^{1023} \pmod{1024}$ is $\dots$
I've tried for the summation like this.
$\sum_{i=1}^{n} (2i-1)^{2i-1} \equiv 0 \pmod{n}$ when $n$ is even. But, I didn't get the answer when $n$ is odd and didn't yet can proved it.
Any hints or idea? Thanks in advanced.
 A: We will prove that for $n\geq 2$ we have
$$
S_n:=\sum_{k=1}^{2^{n-1}}(2k-1)^{2k-1}\equiv 0\pmod{2^n}.
$$
We will induct on $n$ (the case $n=2$ is clear). Consider $S_{n+1}$:
$$
S_{n+1}=\sum_{k=1}^{2^{n}}(2k-1)^{2k-1}=\sum_{k=1}^{2^{n-1}}\left((2k-1)^{2k-1}+(2k-1+2^n)^{2k-1+2^n}\right)\equiv 0\pmod{2^{n+1}}.
$$
Due to Euler's theorem and the fact that $\varphi(2^{n+1})=2^n$ for odd $a$ we have $a^{2^n}\equiv 1\pmod {2^{n+1}}$ (actually $\pmod {2^{n+2}}$ but it's not important now), so
$$
S_{n+1}\equiv\sum_{k=1}^{2^{n-1}}\left((2k-1)^{2k-1}+(2k-1+2^n)^{2k-1}\right)\equiv 0\pmod{2^{n+1}}.
$$
Now, note that because of the Binomial theorem we have ($2^{ns}\equiv 0\pmod {2^{n+1}}$ for $s>1$)
$$
(2k-1+2^n)^{2k-1}\equiv(2k-1)^{2k-1}+\binom{2k-1}{1}(2k-1)^{2k-2}\cdot 2^n\pmod {2^{n+1}},
$$
or
$$
(2k-1+2^n)^{2k-1}\equiv(2^n+1)(2k-1)^{2k-1}\pmod {2^{n+1}}.
$$
Therefore,
$$
S_{n+1}\equiv\sum_{k=1}^{2^{n-1}}\left((2k-1)^{2k-1}+(2^n+1)(2k-1)^{2k-1}\right)=(2^n+2)S_n\equiv 0\pmod{2^{n+1}},
$$
because $2^n+2$ is even and $2^n\mid S_n$ by the inductive assumption.
Thus, $2^n\mid S_n$ for all $n\geq 2$, as desired.
Note. One can extend this argument to prove that map $x\mapsto x^x$ is a bijection on $(\mathbb{Z}/2^n\mathbb{Z})^{\times}$ (just add this to the inductive assumption and prove that $(2k-1)^{2k-1}\not\equiv (2k-1+2^n)^{2k-1+2^n}\pmod{2^{n+1}}$).
