Question

is this a proof that n! > n^2 for n>= 4
So we assume that k^2 < k!
Notice that that k>1 is equivalent to 2k+1 < 3k
It follows that k>= 4 we have
2k+1 < 3k < (k)! k
Now using the assumption
k^2 < k!
we have
k^2 +(2k+1) < k! +(2k+1) + k! +(k)! k=(k+1)!
So
(k+1)^2 < (k+1)!

Answer 1

got no idea, proofs and me are mortal enemies, anything that involves reasoning over common sense is for the birds ;)