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?
 A: In your example you get $2^2\cdot 3$ so the exponents are $2$ and $1$, not three $1$s. This gives $(2+1)\cdot (1+1)=6$, the correct answer.
The way the theorem is proven is by noting that you can choose exactly how many of a specific prime to include. That is, if
$$n=p_1^{e_1}\cdot\ldots\cdot p_r^{e_r}$$
then there are $e_i+1$ choices for how many factors of $p_i$ are in there. $0,1,2,3,4\ldots, e_i$. That's where the factor comes from. If you tried to do, as you do in this instance $2^1\cdot 2^1\cdot 3^1$, then you are saying by putting in the first two $(1+1)$ factors (the ones that go with the $2$s) that you can take $4$ cases:  no $2$s at all, $0$ of the first two and $1$ of the second, $1$ of the first two and $0$ of the second, or $1$ of each $2$. Notice that you count the case of having a single $2$ twice! It doesn't matter which factor of $2$ you grab, it still produces the same factor of $12$. If you are allowed to break up the number, you will always double count something, so you must keep all of the same primes together.
A: 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.
