Here's a proof of the uniqueness of FTA by induction that is different from the proof you cite from Wikipedia. It is due to Zermelo, and isn't as widely known as the other proofs.
Suppose we already proved uniqueness for all numbers $<n$ and are now proving for $n$. If $n$ is prime, there is nothing to prove. Otherwise let $p$ be the smallest divisor of $n$ that is not 1. Clearly $p$ must be prime, otherwise it would not be the smallest.
We claim that any way to write $n$ as a product of primes must include $p$. If we succeed in proving this, our work is done - why? Because if any two factorisations of $n$ must both include $p$, we can take $p$ out, get a smaller number, invoke uniqueness as the inductive assumption and see that the factorisations are really the same.
Consider any factorisation $n=q_1*q_2*\dots*q_s$. If $q_1=p$, we are done. Otherwise we can form a number $n'=p*q_2\dots*q_s$, where we replaced the first factor $q_1$ by $p$. Because $p$ is the smallest divisor, $n'$ must be smaller than $n$. Note that their difference can be written
$n-n' = (q_1-p)*q_2*\dots*q_s$
We know that $p$ divides this number $n-n'$, because it divides both $n$ and $n'$. And this number is less than $n$, so by inductive assumption it only has one factorisation, in which $p$ must participate. But look at the product above: $p$ does not divide $q_1-p$, so the only way it can participate is by being one of $q_2\dots q_s$, which is what we wanted to prove.
Best Answer
Write the fractions $1/n,2/n,3/n \dots ,n/n$ in the simplest form and you can observe that each fraction is of the form $s/t$ where $t$ divides $n$ and $(s,t)=1$. So the number of the fractions is the same as $\sum_{k|n}{\phi(k)}$ which is equal to $n$.