Problem 1597

Proof Let $2=p_{1}<p_{2}<\cdots<p_{n}<\cdots$ be all the prime numbers, then
$$p_{1}, p_{1}^{2} p_{2}, p_{1}^{2} p_{2}^{2} p_{3}, \cdots, p_{1}^{2} p_{2}^{2} \cdots p_{n-1}^{2} p_{n}, \cdots$$

satisfies the requirement.
To verify this claim, we denote the $n$-th number in the sequence as $a_{n}$. First, each $a_{n}$ is not a power. For any $r, s, \cdots, n(1 \leqslant r<s<\cdots<n)$, by (1), $p_{r} \mid a_{r}$ but $p_{r}^{2} \nmid a_{r}$, and $p_{r}\left|\frac{a_{s}}{a_{r}}, \cdots, p_{r}\right| \frac{a_{n}}{a_{r}}$. Therefore, in
$$a_{r}+a_{s}+\cdots+a_{n}=a_{r}\left(\frac{a_{s}}{a_{r}}+\cdots+\frac{a_{n}}{a_{r}}+1\right)$$

the second factor is coprime with $p_{r}$, so the prime $p_{r}$ appears exactly once in the standard factorization of $a_{r}+a_{s}+\cdots+a_{n}$, hence $a_{r}+a_{s}+\cdots+a_{n}$ is not a power. Moreover, since there are infinitely many primes, there are also infinitely many numbers in (1).