Proof by induction, factorials and exponents I've spent many hours trying to find proof by induction to this problem:
$$(n!)^2<\left(\frac{(n+1)(2n+1)}{6}\right)^n$$
for n > 1.
Does anyone have any idea how to solve it? Any help would be appreciated. 

Using AM-GM inequality (which can be proved by induction on the number of terms),
$$\sqrt[n]{1^2\cdot 2^2\cdot3^2 \cdots n^2} \le \frac{1^2 + 2^2+3^2+\cdots + n^2}{n}$$
Equality holds iff $1^2 =  2^2 =  \ldots = n^2$, which means equality does not hold for $n>1$.
On the left hand side,
$$(n!)^2 = 1^2 \cdot2^2 \cdot 3^2 \cdots n^2$$
which can be proved by induction on $n$.
On the right hand side,
$$1^2 + 2^2+3^2+\cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$
which can also be proved by induction on $n$.
Joining the three links together,
$$\sqrt[n]{(n!)^2} < \frac{(n+1)(2n+1)}{6}$$
Taking the $n$th power on both sides (which preserves order as both sides are positive) gives the required inequality.
