I think its clear its not. To begin with for $a, b \in \mathbb{Z}, a^b$ isn't in general an integer. However this came up in a book I was reading discussing RSA public key encryption. The authors make the following argument:
$n|m$ means n divides m, for $m \in \mathbb{Z}$, let $[m]_n = \{x \in \mathbb{Z} : n|(x-m)\}$, $gcd(m, n)$ is the greatest common denominator of $m$ and $n$, and $\phi(n)$ is Euler's $\phi$-function.
Let $m$, $n$, and $e$ be positive integers, and let $n > 1$, $gcd(m, n)=1$, and $gcd(e, \phi(n))=1$. Let $d$ be the smallest possible integer such that $d\equiv k\ (mod\ \phi(n))$. Since $gcd(e, \phi(n))=1$, there are two integers $k$ and $l$ such that $1 = ke + l\phi(n)$, and since $d\equiv k\ (mod\ \phi(n))$, $1 = de + j\phi(n)$ for some integer $j$.
Now since $gcd(m, n)=1$, by Euler's theorem, $m^{e^{d}}=m^{1-j\phi(n)}=m\cdot m^{\phi(n)^{-j}}\equiv m \cdot 1 \equiv m$.
But what if $j\geq0?$ For some $b \in \mathbb{Z}^+$, I think $([m]_n)^b$ makes sense, since $[m_1]_n\cdot[m_2]_n=[m_1m_2]_n$, so $([m]_n)^b=([m^b]_n)$. And then since congruence modulo n is an equivalence relation, if by Euler's theorem $a^{\phi(n)} \equiv 1\ (mod\ n)$, then $[a^{\phi(n)}]_n = [1]_n$ and $([a^{\phi(n)}]_n)^b = ([1]_n)^b$. However this doesn't work if b is $0$ or a negative integer. Is it enough to say that $1 = 1^b$ for all $b\in \mathbb{Z}$, positive or not, and make the same argument I just made?
