I assume that by $n$, you mean a natural number $n\in\mathbb N$.

If $|\mathbb S|=m$ is finite, this is simply a matter of combinatorics: for each coordinate in the $n$-tuple $(x_1,\dots,x_n)$ we have $m$ options, so the total number of $n$-tuples is simply $n\cdot m$.

If $\mathbb S$ is infinite, things get a little more complicated. For $\aleph$ numbers we have that $\aleph_\alpha\cdot\aleph_\alpha=\aleph_\alpha$. You can prove this by using the canonical well-order on $Ord^2$ by letting $(\alpha,\beta)<(\gamma,\delta)$ iff

- $\max\{\alpha,\beta\}<\max\{\gamma,\delta\}$, or
- $\max\{\alpha,\beta\}=\max\{\gamma,\delta\}$ and $(\alpha,\beta)$ is less than $(\gamma,\delta)$ using the lexicographical order.

If there exists a bijection between $\mathbb S$ and an aleph number $\aleph_\alpha$, then we can see that $\mathbb S^n$ has cardinality $|\mathbb (\aleph_\alpha)^n|=\prod_n \aleph_\alpha=\aleph_\alpha$. When the Axiom of Choice is true, any infinite $\mathbb S$ has a cardinality equal to an aleph number.

However, without choice, there exist sets $\mathbb S$ that have a cardinality unequal to any $\aleph$ number, and hence we cannot say $\mathbb S^n$ has the same cardinality as $\mathbb S$. In fact without the Axiom of Choice it is consistent that $|\mathbb S|\neq |\mathbb S\times\mathbb S|$. The statement that $\mathbb S$ and $\mathbb S\times \mathbb S$ have the same cardinality is sufficient to prove the Axiom of Choice.

With or without choice, you could define the cardinality of $\mathbb S^n$ as the cardinality of the set of functions $f:\{0,1,\dots,n-1\}\to\mathbb S$. This is how $|\mathbb S^n|$ is defined. Since $n=\{0,1,\dots,n-1\}$ is a cardinal itself, and cardinal exponentiation $\kappa^\lambda$ is defined as the cardinality of the set of functions $\lambda\to\kappa$, we see that $|\mathbb S^n|=|\mathbb S|^n$.