# A proof using the phi-function

$\sqrt{n} < \phi(n) < n-\sqrt{n}$
So, given an arbitrary positive integer $k$, $\phi(n)>k$ for all $n>k^2$. So the highest number $n$ satisfying $\phi(n)=k$ must still be less than $k^2$. Since there is such an upper bound, the total number of integers n satisfying this equality is finite.