Beginner Mathematics-3

The problem appears to be only considering $n\geq8$ . For $n<8$ , there is a solution $n=4$ , which yields $n!=24$ .

For $n\geq 8$ , we can rewrite the expression as $2^8(1+2^2+2^{n-8})=2^8(5+2^{n-8})$ . This will be a perfect square precisely when the expression in the parentheses is a perfect square. Let $m=n-8$ . Note that $5+2^m$ is odd (unless $m=0$ , in which case we get $5+1=6$ , which is not a square). Therefore, if it is to be a perfect square, it must be congruent to $1\bmod 8$ . Since $5\equiv 5\bmod 8$ , we must have $2^m\equiv 4\bmod 8$ . The only power of $2$ that is congruent to $4\bmod 8$ is $2^2=4$ . Thus $m=2$ and $n=10$ , and our solution is $10!=\boxed{3628800}$ .

$(2^4+2^5)^2 = 2^8 + 2(2^4)(2^5) + 2^{10} = 2^8 + 2^{10} + 2^{10}$ $\implies n = 10$ and $n! = 10! = \boxed{3628800}$ .