Let $M=2\times 3^2\times 5^3\times 7^4$. Let $a_1,a_2,...,a_{120}$ be the positive divisors of $M$. Calculate $\prod_{i=1}^{120} a_i$ Let $M=2\times 3^2\times 5^3\times 7^4$. Let $a_1,a_2,...,a_{120}$ be the positive divisors of $M$. Calculate $$\prod_{i=1}^{120} a_i$$.
My Attempt:
I am aware how to find the sum of these $120$ divisors.$$\sum_{i=1}^{120}a_i=(2^0+2^1)(3^0+3^1+3^2)(5^0+5^1+5^2+5^3)(7^0+7^1+7^2+7^3+7^4)$$But I have never come across a problem asking for product of divisors before.
In the product of divisors $2^0$  and $2^1$ will occur $(1+2)(1+3)(1+4)=60$ times each.
Similarly, $3^0$,$3^1$ and $3^2$ will occur $(1+1)(1+3)(1+4)=40$ times each.
$5^0, 5^1, 5^2, 5^3$ will occur $(1+1)(1+2)(1+4)=30$ times each.
$7^0, 7^1, 7^2, 7^3, 7^4$ will occur $(1+1)(1+2)(1+3)=24$ times each.
So product of divisors equals $$(2^0)^{60}.(2^2)^{60}.(3^0)^{40}.(3^1)^{40}.(3^2)^{40}.(5^0)^{30}.(5^1)^{30}.(5^2)^{30}.(5^3)^{30}.(7^0)^{24}.(7^1)^{24}.(7^2)^{24}.(7^3)^{24}.(7^4)^{24}=2^{60}.3^{120}.5^{180}.7^{240}$$
Is this correct.
Can there be a shorter or more direct way to do it.

Let's order the $a_{1}, \ldots, a_{120}$ so that $a_{i} \cdot a_{121-i} = M$ for all $1 \leq i \leq 120.$ We can do this because an integer $d$ is a positive divisor of $M$ if and only if $\frac{M}{d}$ is a positive integer divisor of $M$ (and since $M$ is not a square, $d$ and $\frac{M}{d}$ are never equal).
Now, note that
$$\left(\prod_{i=1}^{120}a_{i}\right)^{2} = \left(\prod_{i=1}^{120}a_{i}\right)\left(\prod_{i=1}^{120}a_{121-i}\right),$$ because the second product on the right side is just a permutation of the first, and multiplication is commutative.
Then, we can write
$$\left(\prod_{i=1}^{120}a_{i}\right)\left(\prod_{i=1}^{120}a_{121-i}\right) = \prod_{i=1}^{120} a_{i}a_{121-i},$$ once again using the commutativity of multiplication. However, since $a_{i}a_{121-i} = M$ for all $i$, it follows that
$$\prod_{i=1}^{120} a_{i}a_{121-i} = \prod_{i=1}^{120} M = M^{120}.$$
So, since $$\left(\prod_{i=1}^{120}a_{i}\right)^{2} = M^{120},$$ we see that
$$\prod_{i=1}^{120}a_{i} = M^{60}.$$
Finally, since $M = 2 \times 3^{2} \times 5^{3} \times 7^{4}$, we have
$$\prod_{i=1}^{120}a_{i} = 2^{60} \times 3^{120} \times 5^{180} \times 7^{240},$$
which agrees with what you got.
