Does $ \sum_{i=1}^n n^{k_i} $ determines $(k_1,...,k_n)$? Let $k_1,...,k_n\in\mathbb{N}$. Does the power sum
$$
\sum_{i=1}^n n^{k_i}
$$
uniquely determines the $n$-tuple $(k_1,...,k_n)$?
Remark: In the case $n=2$, this is true. However, when trying to generalize to an arbitrarily sized sum, it doesn't hold. For example,
$$
2^0+2^0+2^2=2^1+2^1+2^1.
$$
I then thought of a fixed sum size, equal to the considered base, but I don't know exactly how to argue or build this bijection. The motivation behind this question comes from trying to determine the $n$-tuple $(k_1,...,k_n)$ from the sum
$$
\sum_{i=1}^n g(k_i),
$$
where $g$ is some constructible function.

Obviously you can only hope to determine the $n$-tuple up to permutation, and you can only hope to determine even that when $n>1$. So I'll assume that's what you want. And as you see, it doesn't work in general when the number of $k_i$ is not equal to the base of the exponent.
Then the answer is yes, if you know $n$. It's a corollary of the uniqueness of base-$n$ representations. You can read off the $k_i$ from the digits of the number $s = \sum_{i=1}^n n^{k_i}$ written base $n$. The $j^{\small\text{th}}$ digit ($j=0$ is the first digit) will tell you how many $k_i$ are equal to $j$, unless there is exactly one $1$ and no other nonzero digits in $s$, in which case all the $k_i$ are equal to $j-1$ where $j$ is the place where the $1$ occurs.
