Here is a thought experiment: Suppose we did not know what sets and functions are. The general idea of a topos is, that it somehow serves as a foundation for mathematics. So let there be an alternate world and a well known topos $\mathcal{A}$, "in which everyone does mathematics", in this world. Eventually, some category theorist claims: "Oh, our mathematical universe is not as 'fundamental' as we think. I found a topos that I henceforth shall call $\mathsf{Set}$ and an $n\in \mathbb{N}$ (not $1$) such that $\mathcal{A}$ is equivalent to $\mathsf{Set}^n$" (because $\mathcal{A}$ was $\mathsf{Set}^{45634}$ all along, for example).
Let us come back to this world with the obvious question:
Could there be a category $\sqrt[n]{\mathcal{\mathsf{Set}}}$ and an $n\in \mathbb{N}, n>1$ with $\sqrt[n]{\mathcal{\mathsf{Set}}}^n $ equivalent to $\mathsf{Set}$?
What if instead the $n$ is any small category besides $1$?
There might be an issue with this question: I am not sure, what "category" really should mean in this context. Possibly, the "right" answer assumes, that category theory deals with categories internal to $\sqrt[n]{\mathcal{\mathsf{Set}}}$. I do not know that, since I am no expert by any means.
It is not possible.
Let $\mathcal{C}$ be a category. First things first: by a symmetry argument, we see that the diagonal of $\mathcal{C}^n$ is closed under whatever limits or colimits exist in $\mathcal{C}$; in particular, if $\mathcal{C}^n$ has finite limits and colimits, then so does $\mathcal{C}$. Suppose $\mathcal{C}$ has a terminal object $1$ and the binary coproduct $1 + 1$. Then, $$\mathcal{C}^n ((1, \ldots, 1), (1 + 1, \ldots, 1 + 1)) \cong \mathcal{C} (1, 1 + 1)^n$$ and in particular, if this set is finite, then the number of elements is an $n$-th power. But if $\mathcal{C}^n$ is equivalent to $\mathbf{Set}$, this forces $n = 1$.