What is the integral part of *M*, if *M*= $\sqrt{2012* \sqrt{2013* \sqrt{2014* \sqrt{... \sqrt{( 2012^2-2)* \sqrt{(2012^2-1)* \sqrt{2012^2}}}}}}}$ What is the integral part of M, if 
M=
$\sqrt{2012* \sqrt{2013* \sqrt{2014* \sqrt{... \sqrt{( 2012^2-3)* \sqrt{(2012^2-1)* \sqrt{2012^2}}}}}}}$
I have no clue how to solve this, hints aswell as full solutions would be appreciated, taken from the 2012 IWYMIC https://chiuchang.org/wp-content/uploads/sites/2/2018/01/2012-IWYMIC-Individual.x17381.pdf

For convenience let $$f(n)=\sqrt{n\sqrt{(n+1)\sqrt{(n+2)\cdots\sqrt{(n^2-2)\sqrt{(n^2-1)\sqrt{n^2}}}}}}$$
where $n$ is a positive integer.  So basically,
$$f(n)=\prod_{k=n}^{n^2}k^{\frac1{2^{k-n+1}}}.$$
Let $n>1$.  As $k> n$ for each index $k=n+1,n+2,\ldots,n^2-2,n^2-1$, we get 
$$f(n)>n^{\frac{1}{2}+\frac1{2^2}+\ldots+\frac{1}{2^{n^2-n}}}\cdot (n^2)^{\frac{1}{2^{n^2-n+1}}}=n^{1-\frac{1}{2^{n^2-n}}}n^{\frac1{2^{n^2-n}}}=n.$$
For the upper bound, we can easily see that
$$\sqrt{n^2}<n^2+1.$$
So $$\sqrt{(n^2-1)\sqrt{n^2}}<\sqrt{(n^2-1)(n^2+1)}<\sqrt{n^4}=n^2.$$
We continue by induction to show that
$$\sqrt{k\sqrt{(k+1)\sqrt{(k+2)\cdots\sqrt{(n^2-2)\sqrt{(n^2-1)\sqrt{n^2}}}}}}<k+1$$
for all positive integers $k\leq n^2$.  In particular this shows that
$$f(n)<n+1.$$
Therefore
$$n\leq f(n)<n+1$$
for every positive integer $n$, with equality on LHS iff $n=1$.
