jred wrote: Find all positive integers $n$ such that there exists integers $n_1,\ldots,n_k\ge 3$, for some integer $k$, satisfying \[n=n_1n_2\cdots n_k=2^{\dfrac{1}{2^k}(n_1-1)\cdots (n_k-1)}-1.\] All $n_1,n_2,...,n_k$ are odd since they are factors of $2^p-1$ where $p=\dfrac{1}{2^k}(n_1-1)\cdots (n_k-1)$. Hence $n_i\geq5$. For all $p\geq10$, we have $2^p-1> p^3$. And for $n_i\geq5$, we see that $\left(\dfrac{n_i-1}2\right)^3>n_i$ yielding a contradiction \[2^p-1>n_1n_2\cdots n_k=n.\] So, $p<10$ and we see that only $p=3$ fits.

Why do you assume $n_i\geq5$?

I think the original problem says $n_1,n_2,...,n_k>3$? In that case, since $n_i$ are odd, we've $n_i\geq 5$.