# Sum Equals Product: A Diophantine Equation

I formulated the following claim after reading Problem 88 of Project Euler:

Fix $k$ and let $\mathscr N$ be the set of numbers $N$ satisfying $$N=n_1+n_2+\cdots+n_k=n_1n_2\cdots n_k,$$ where each $n_i$ is a positive integer.

Claim: $\mathscr N$ is finite for every $k>1$.

I believe that it is true, but I do not know how to prove it.

On a side note, I proved that $\mathscr N$ is nonempty for every $k>1$:

\begin{align}N=2+k+\overbrace{1+1+\cdots+1}^{k-2\text{ times}}=2k&\implies2+k+k-2=2k\\&\implies2k=2k\\&\implies N\in\mathscr N.\end{align}

Could it be that $2k$ bounds $N$?

Wlog. $n_1\ge n_2\ge\ldots \ge n_k$. We cannot have $n_1=1$ as that would lead to $k\cdot 1=1^k$: Let $m$ be the largest index with $n_m>1$. We have $m=1$ would lead to $n_1+k-1=n_1$, so we must have $m\ge 2$. Then, supposing $n_1>k$, $$n_1+\ldots+n_m+1+\ldots +1\le mn_1+k-m<n_1(m+1)-1$$ $$n_1\cdots n_m\cdot 1\cdots 1\ge n_12^{m-1}$$ But then $2^{m-1}<m+1$. Conlcude $m\le 2$ and continue from there