Sudheesh Singanamalla
Last Activity: 14 Years ago
given : n^n
to prove : n^n = 2n!
for n=1
(1^1 = 1) <= (2*1 = 1)
since it is true for n=1, for n=k
k^k <= 2*k!
assuming n=k to be true, multiplying bothsides by k we get
k^k * k^1 <= (2*k)! * k
k^k+1 <= (2*k*k) * (2*(k-1))! = 2[(k*k)*(k-1)]!
this is of the form n^n = 2n!
therefore n^n = 2n!