The general form of this kind of number is $2^{15}3^{10}5^{6}x^{30}$, for all values of $x$.
When $x=1$, this number is the product of powers of 2, 3, and 5 just under some integer, eg $32768$, $59049$ and $15625$ are the largest powers under, say $60,000$.
The general working for such a number is to treat the indeces by chinese remainder. So, for example, should we seek something where $N/2$ is square, $N/3$ is cube, and $N/5$ is a fifth power, we would note that, eg it is a product of some powers of 2, 3, and 5, eg $2^a 3^b 5^c$.
We have then that 2 divides $a-1, b, c$, 3 divides each of $a, b-1, c$ and 5 divides each of $a, b, c-1$. We get then that 15 divides a, 10 divides b, and 6 divides c.
- We find now a multiple of 15 that 2 divides a-1: result = $a=15$
- We find then that 10 that 3 divides b-1, result $b = 10$
- We find a result where 5 divides c-1: result = $6$.
We then note that every number that is a power of 30, is already a square, a cube, and a fifth power, ie $x^{2 \cdot 3 \cdot 5} = x^{30}$, and thus if $N/2$ is square, so is $(N\cdot x^{30})/2$ is also square, and if $N/3$ is a cube, so is $(N\cdot x^{30})/3$, and if $N/5$ is a fifth power, so is $(N\cdot x^{30})/5$, is true for all $x$.
In the worked example, we include a final row, for all $x$, where the power of $x$ is a multiple of 2, 3, 5, and 7, and therefore satisfies all of the conditions above.
If we wish to further add $N/7$ as a seventh power, then we would need to look for some set where $2 | (a-1, b, c, d) $, $3 | (a, b-1, c, d)$, etc. This means that eg $b, c, d$ are even, $a, c, d$ are multiples of 3, etc. Here, we write the modulus in list form, where the modulus is taken over the corresponding point in the mod function.
We see that in the first row, $a$ must give a remainder of 0, when dividided by 3, 5, 7, so it must be a multiple of the product of $3 \cdot 5 \cdot 7$, that leaves a remainder of $1$, when divided by $2$.
- a = 1, 0, 0, 0, mod(2, 3, 5, 7), is a multiple of 105, gives $a=105$
- b = 0, 1, 0, 0, mod(2, 3, 5, 7), is a multiple of 70, gives $b=70$
- c = 0, 0, 1, 0, mod(2, 3, 5, 7), is a multiple of 42, gives $c=126$
- d = 0, 0, 0, 1, mod(2, 3, 5, 7), is a multiple of 30, gives $d=120$
- e = 0, 0, 0, 0, mod(2, 3, 5, 7), is a multiple of 210, gives $e=210$
So we can write this as $2^{105} 3^{70} 5^{126} 7^{120} x^{210}$, for all values of $x$. Note that we don't need to write $x^{210y}$, since the range of $x$ already includes $x^y$.
Best Answer
If you factor a perfect square into primes, every prime has an even exponent. The argument made says that for all $n \gt 5,$ there are at least two factors of $2$ in $n!$. In fact, this is true for all $n \ge 4$, but they covered $4$ and $5$ in the casework. Once $n!$ is a multiple of $4$, $n!+10\equiv 2 \pmod 4$ has an odd exponent of $2$, so it cannot be a perfect square.