Show that $f(g(x)) = x$ and $g(f(x))= x$, with $f(x) = x^e \bmod n$ and $g(x) = x^d \bmod n$ I want to solve the following problem:

Let $d$ and $e$, both natural numbers, be each others inverses modulo $\varphi(n)$, where $n = p\cdot q$ is a product of two different prime numbers $p$ and $q$. Let $M = \{0,1,2,\dots,(n-1)\}$ be the set of nonnegative numbers smaller than $n$. Define two functions $f: M \rightarrow M$ and $g: M \rightarrow M$ as
  \begin{align*}
f(x) = x^e \bmod n \quad \mbox{and}\quad g(x) = x^d \bmod n
\end{align*}
  Show that $f(g(x)) = x$ and $g(f(x))= x$ for all $x \in M$. 

I understand that $f(x)$ and $g(x)$ will always produce numbers between 0 and $n$, since $x$ is smaller than $n$. In that respect, $f(x) = g(x)$ no matter what $e$ and $d$ we choose.
But I don't understand why $f(g(x)) = x$ and $g(f(x))= x$. 
 A: One has $ed=k\varphi(n)+1$ for some integer $k$ because $ed\equiv 1\pmod{n}$. So if $\gcd(x,n)=1$, we can write bearing in mind that $x^{\varphi(n)}\equiv 1\pmod{n}$
$$f(g(x))=x^{ed}=x^{k\varphi(n)+1}=\left(x^{\varphi(n)}\right)^k\cdot x^1=x$$
Similarly we prove that $g(f(x))=x$
When $\gcd(x,n)\gt 1$ it doesn't work as shown by the example $n=4$, $x=2$, $e=1$ and $d=3$.
A: I think I figured it out. First I have to prove that $x^{k\varphi(n) + 1} \equiv x \pmod{n}$, even when I don't know if $\gcd(x,n) = 1$.    
We look at the system 
\begin{align}
\begin{cases}
y \equiv x \pmod p \\
y \equiv x \pmod q
\end{cases}
\end{align}
Since $q$ and $p$ are two different prime numbers, they are relatively prime to eachother. Then we have
\begin{align*}
&\varphi(n) = \varphi(p)\cdot \varphi(q)
\end{align*}
and so, by Eulers theorem, 
\begin{align*}
&x^{k\varphi(n) + 1} = (x^{\varphi(p)})^{k\varphi(q)} \cdot x \equiv 1^{k\varphi(q)} x \equiv x \pmod{p}\\
&x^{k\varphi(n) + 1} = (x^{\varphi(q)})^{k\varphi(p)} \cdot x \equiv 1^{k\varphi(p)} x \equiv x \pmod{q}
\end{align*}
Thus, a solutions to the set of congruences above is 
\begin{align*}
y =  x^{k\varphi(n) + 1}
\end{align*}
By the Chinese Remainder Theorem, this solution is unique modulo $p\cdot q =n$. Thus, 
\begin{align*}
x^{k\varphi(n) + 1} \equiv x \pmod{n}
\end{align*}
Then, I can apply the solution as proposed by marwalix, namely
\begin{align*}
&f(g(x)) = x^{ed} = x^{k\varphi(n) + 1} \equiv x \pmod{n}\\
&g(f(x)) = x^{de} = x^{k\varphi(n) + 1} \equiv x \pmod{n}
\end{align*}
