What are the notations $k^{\prime n}$ and $\varphi^n$ in algebra?

Question

I would like to understand what the following problem says:

Let $k$ be a commutative ring and $f\in k[X_1,\ldots,X_n]$. Let $k^\prime,k^{\prime\prime}$ be commutative $k$-algebras and $\varphi:k^\prime\to k^{\prime\prime}$ be a $k$-homomorphism. Prove that for every $x\in k^{\prime n}$ we have $$f(\varphi^n(x^\prime))=\varphi (f(x^\prime))$$

Is $k^{\prime n}=(k^\prime)^n=k^\prime \times \cdots\times k^\prime$ and $\varphi^n(x^\prime)=(\varphi (x^\prime))^n$ or $\varphi^n(x^\prime)=\varphi(\varphi (\ldots \varphi(x^\prime)\ldots ))$?

This is from http://www.math.helsinki.fi/kurssit/alggeom/h1.gif problem 3.

Answer 1

Since $f$ is a multivariate polynomial, $\varphi^n(x')$ must be a list of $n$ elements of some ring. The notation is unclear, but based on the context of what I understand from the problem $\varphi^n(x')$ is $$(\varphi(x_1'),\varphi(x_2'),\ldots,\varphi(x_n'))$$ where $x_i'$ are the components of $x'$, and $(k')^n$ is $k'\times k'\times\cdots\times k'$ ($n$ times).

Answer 2

It is the case $k^{\prime n}=(k^\prime)^n=k^\prime \times \cdots\times k^\prime$.

Yet $\varphi^n$ is neither $\varphi^n(x^\prime)=(\varphi (x^\prime))^n$ nor $\varphi^n(x^\prime)=\varphi(\varphi (\ldots \varphi(x^\prime)\ldots ))$.

The map $\varphi^n$ is a map defined on $k^{\prime n}$ so $n$-tuples; where in each coordinate you use $\varphi$.