How did this method get the correct answer? Product of divisors Find the product of the divisors of $420^4$.
$420^4=2^8 \cdot 3^4 \cdot 5^4 \cdot 7^4$. I tried the following method which I soon realized was flawed: for $2^8$ there are $5^3$ possible divisors, for $2^7$ there are also $5^3$ and so on... for $3^4$ there are $9 \cdot 5^2$ possible divisors and for $3^3$ and so on and so forth.
Though I seem to be over-counting the powers loads here, e.g. I will count any one divisor maybe four times. But I still get the answer as $420^{2250}$ which is correct, how can this be?
And yes I know the formula for working out the product easily.
 A: I believe it's just a coincidence of the fact that you're computing the product of the divisors of $n^{\ell}$ for a number $n$ that has $\ell$ distinct prime factors.  
Set $n=420$ and $m=420^4$.  If by the statement "for $2^8$ there are $5^3$ possible divisors" you mean that there are $5^3$ possible divisors of $m$ of the form $2^8 \cdot s$ for an integer $s$, then yes, this will over-count the number of divisors $d$ of $m$ by a factor of $4$.  So your count is $c = 4d$.
I suspect all you did was compute $n^{c/2}$ rather than $m^{d/2}$.  But $n^{c/2} = n^{4d/2} = (n^4)^{d/2} = m^{d/2}$.  
This generalizes.  If $n$ has the prime factorization $n=\prod_{i=1}^{\ell} p_i^{a_i}$ then
$$
c:=\prod_{i=1}^{\ell} (a_i + 1) = \ell \cdot d
$$
where $d$ is the number of divisors of $n$.  Then of course the product of all the divisors of $n^{\ell}$ is just $n^{c/2}$.  
A: What you are doing seems to be correct, although your procedure is a bit ad hoc.
Let me try to formalize your method: in general, if $x = p_1^{n_1} p_2^{n_2} ... p_k^{n_k}$ with $p_i$ distinct primes and $n_i>0$, the product of all factors will be of the form $p_1^{K_1}\dots p_n^{K_n}$ for some $K_i >0$. The crux of the matter is to determine the $K_i$'s. Note that $K_i$ will be the sum over all factors of $x$ of the exponent of $p_i$ appearing in the factor. 
This is, I think, what you are getting at when you write "for $2^8$ there are $5^3$...". You're saying that factors divisible by $2^8$ will contribute $8\cdot 5^3$ to the exponent of $2$ in the product of all factors. Factors divisible by $2^7$ but not $2^8$ will contribute $7*5^3$, and so forth. Summing all these, you get $(1+2+3+4+...+8)(5^3)$ for the exponent of $2$. The way that you've enumerated things, nothing is double counted.
Returning to the general set up, factors of $x$ are precisely numbers of the form $p_1^mt$, for $t$ a factor of $p_2^{n_2}\dots p_k^{n_k}$ and $0 \leq m \leq n_1$. If we let $M$ equal the number of factors of $p_2^{n_2}\dots p_k^{n_k}$, we have that $$K_1=M(\sum_{i=1}^{n_1}i) = M\frac{n_1(n_1+1)}{2}.$$ In your case, if $p_1=2$ and $n_1=8$, you calculate that $M= 5^3$. This is correct: in general, $M$, the number of factors of $p_2^{n_2}\dots p_k^{n_k}$, is $(n_2+1)\dots (n_k+1)$. Generalizing this argument to other indices, 
$$K_j = \frac{(\prod_{i=1}^k n_i+1)(n_j)}{2},$$
which is the formula that you know. So your method can actually be used to construct a (much longer) proof of the formula.
Hope this helps clarify things.
A: Call $n=420^4$, $D = \{ \mbox{divisors of }n\}$ and $d=\#D$.
Then $$
\left(\prod_{a \in D} a\right)^2 = \prod_{a \in D} a * \prod_{a \in D} \frac{n}{a} = n^d
$$
So the answer is $n^{d/2}$.
Finally, computing $d=9*5*5*5 = 1125$, you get that the product of the divisors of $n$ is $420^{2250}$.
