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you'll have to interpret my quick explanation, but it's the following:
change n to n-1, left becomes (n-1)!, right becomes (n^(n-1))/(2^(n-1))
simplify right, it becomes (2*n^n)/(n(2^n)), then multiply bottom to other side.
on the left the n makes the (n-1)! into n! again and the number of 2s is the same as the number of terms in factorial, so each term gets one, and the last term, the 1, becomes a 2. Use it to cancel the 2 on right side.
then use the 2 in you factorial to make it 2^n on the left again. if you followed me, you'll be left with 2^n on the left and n-2 terms of the fact. and on the right you'll have n terms of n, so for example if n was 7 it would be (2^7)*7*6*5*4*3 < 7*7*7*7*7*7*7
divide all the 2s over except 1 and cancel an n and you have 6,5,4,3,2 versus (3.5)^6
which is your k-1 case so you can show
"Assume that k! < ((k+1)/2)^k holds for k > 1
We must show that (k+1)! < ((k+1 + 1)/2)^(k+1)"
something like that, sorry it's a mess, but there's your algebra, i should be working in something else.....
