math question of the day:
Prove that
$$\forall n \in \mathbb{N} (n \ge 1 \rightarrow n^{n} \ge n)$$

Question

math question of the day:
Prove that
$$\forall n \in \mathbb{N} (n \ge 1 \rightarrow n^{n} \ge n)$$

anonymous · Answer

oopes thats $$n^{n} \ge n!$$

hoblos · Answer

i think it can be solved by induction....

hoblos · Answer

for n=1
1 >=1   true

hoblos · Answer

if we suppose it is true for n 
n^n  >= n!

nenadmatematika · Answer

then we proove that is correct for n+1 so (n+1)^(n+1)>=(n+1)!  (n+1)^n*(n+1)>=(n+1)*n! after cancelling n+1 we get (n+1)^n>=n! so our assumption was that n^n>=n! so we proved it...is this OK hoblos ?

hoblos · Answer

for n+1
(n+1)^(n+1) = (n+1)(n+1)^n

but we know that (n+1)>n
                         (n+1)^n >n^n    since n>=1
then
 (n+1)^(n+1) = (n+1)(n+1)^n > (n+1)n^n
                                             > (n+1)*n!
                                             > (n+1)!

hoblos · Answer

@nenadmatematika i think your way is right  too