Factorial

Well, there are two ways of solving this. One way is to rearrange the given value in the form of a factorial and keep count of the prime factors to get, $n!=16! \implies \boxed{n=16}$

Another way is to use reasoning. We see that there are $15$ 2's in the expansion, so using the method of finding trailing zeros, we can also find that if there are $15$ 2's in the expansion, then we have $n\in \{16,17\}$ since if $n\lt 16$ , there will be less number of 2's and if $n\gt 17$ , there will be more number of 2's. Also, observe that $17$ is a prime number, so if $n=17$ , there should be a $17$ in the given prime factorization. But there isn't! So, we have the result $n=16$ assuming the given factorial exists. We can verify the assumption easily.

Hence, the answer is $\boxed{n=16}$ $_\square$

2^15 tell us that n must have 15 factors 2, then the unique possible value of n is 16 because there are 8+4+2+1 factors 2. (Because 8 terms of 16! are divisibly by 2 once, but 4 of the latter are divisibly by 2 twice..ecc.)