<s>[INST] Contradiction on prime decomposition Take $n = 12$
$12$'s prime factorization is $2^1\times2^1\times3^1$
So then, the number of factors by UFT is $(1+1)(1+1)(1+1) = 8$
But there's only $1,2,3,4,6,12 = 6$ factors!!
Where are the other two?
 [/INST] The formula you've written assumes that all factors in the product are distinct. The reason for the product of $(e+1)$ over each factor $p^e$ is because we are multiplying $p^k$ for all $0\le k\le e$, for each factor $p^e$ in the product. Thus, the decomposition $2^12^13^1$ as you've given gives the following $8$ factors:
\begin{align}
2^02^03^0&=1\\
2^12^03^0&=2\\
2^02^13^0&=2\\
2^12^13^0&=4\\
2^02^03^1&=3\\
2^12^03^1&=6\\
2^02^13^1&=6\\
2^12^13^1&=12
\end{align}
And now you can see that the double-counted factors come from the fact that $2^12^0=2^02^1$. As long as the prime bases are different, this can't happen, which is why we demand that the decomposition use distinct prime factors in the factorization that yields that formula.
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