Let n be a positive integer such that $n>1$. Show that for any positive a, $a^{n}\cong a^{n-\phi(n)}$ mod n.
I think that this needs the fact that if positive integers x,y with $y>1$ then $y^{x-1}$ is greater than or equal to x.
Any suggestions? Thanks!
$$a^n\equiv a^{n-\phi(n)}\pmod{n}\iff a^{\phi(n)+n}\equiv a^n \pmod{n}\iff a^{\phi(n)}\equiv 1\pmod{\frac{n}{GCD(n,a^n)}}$$ (e.g. $5^4\equiv 1\pmod{2}$ with $4=\phi(10)$ and $2=10/5$).
If you've proven that $\phi$ is weakly multiplicative, then you'll know that $\phi\left(\frac{n}{GCD(n,a^n)}\right)|\phi(n)$. Thus, this problem reduces to proving that $a^{\phi(n)}\equiv 1\pmod{n}$ if $n$ is relatively prime to $a$. Now, consider all $a_i$ in the unit group of $n$. Note that the map $a\cdot a_i$ is bijective, since $a$ has a multiplicative inverse $\pmod{n}$. Thus, since $\{a\cdot a_i\mid a_i<n,\ GCD(a_i,n)=1\}$ is the same set as $\{a_i\mid a_i<n,\ GCD(a_i,n)=1\}$.
$$\begin{align} \Longrightarrow \prod_{i=1}^{\phi(n)} a\cdot a_i&\equiv\prod_{i=1}^{\phi(n)} a_i \pmod{n}\\ a^{\phi(n)}\prod_{i=1}^{\phi(n)} a_i&\equiv\prod_{i=1}^{\phi(n)} a_i \pmod{n} \\ a^{\phi(n)}&\equiv 1\pmod{n} \end{align}$$
