# 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.