Induct on $n$. In the base case $n = 2$, $1^1 + 3^3 = 28$ is a multiple of $4$ but not $8$. Now suppose $n \geq 3$ and assume this is true for $n - 1$, so \[1^1 + 3^3 + \cdots + (2^{n - 1} - 1)^{2^{n - 1} - 1} \equiv 2^{n - 1} \pmod{2^n}.\]Now note that for all odd $x$, we have \[\nu_2(x^{2^{n - 1}} - 1) = \nu_2(x^2 - 1) + \nu_2(2^{n - 2}) \geq n + 1,\]so then $x^{2^{n - 1}}$ is always $1$ mod $2^{n + 1}$. So then \[(x + 2^{n - 1})^{x + 2^{n - 1}} \equiv (x + 2^{n - 1})^x \equiv x^x + x^x \cdot 2^{n - 1},\]since $2(n - 1) \geq n + 1$. So this means \[1^1 + 3^3 + \cdots + (2^n - 1)^{2^n - 1} \equiv (2 + 2^{n - 1})(1^1 + 3^3 + \cdots + (2^{n - 1} - 1)^{2^{n - 1} - 1}) \pmod{2^{n + 1}}.\]But $2 + 2^{n - 1}$ is even, and the second sum is $2^{n - 1}$ mod $2^n$ by the induction hypothesis, so then the original sum is $2^n$ mod $2^{n + 1}$, as desired.