Find the integral part of $S$. It is given that $S=\sqrt{2012\sqrt{2013\sqrt{2014\sqrt{\cdots \sqrt{(2012^2-2)\sqrt{(2012^2-1)\sqrt{2012^2}}}}}}}$. Find the integral part of $S$.
Thanks in advance!
 A: First, by numeric calculation, it is easy to find that S>2012, we only have to calculate some very small number of terms to get the result.
But for S<2013, it is hard to show that even with Mathematica, it will take a  very long time.
I cannot find a good method, but I checked that your conclusion is right.
$$log(S)=\Sigma_{i=0}^{2012^2-2012}\frac{1}{2^{i+1}}log(2012+i)$$.
Then we can do the sum for the first few N=10000 terms with Mathematica,.Then for the other term with i larger than 10000, we can approximate the sum with integral:
$$\Sigma_{i=10000}^{2012^2-2012}\frac{1}{2^{i+1}}log(2012+i)\approx \int_{10000}^{2012^2-2012}dx \frac{1}{2^{x+1}}log(x+2012)$$ which is quite small. It's much smaller than $log(2013)$ minus the  sum from 0 to N, thus your conclusion is proved to be true.
A: This is a partial answer!

Define
$$
S(n) := \sqrt{n \sqrt{(n+1) \dotsm \sqrt{n^2}}} = \prod_{k=0}^{n(n-1)} (n+k)^{2^{-k-1}}.
$$
So far I can only prove that
$$
n^2 \leq S(n)^2
$$
for every $n \geq 1$. Indeed, first observe that this is trivial for $n = 1$, so let's assume $n \geq 2$. Then note that
$$
S(n)^2 = \prod_{k=0}^{n(n-1)} (n+k)^{2^{-k}} \geq n \prod_{k=1}^{n(n-1)} (n+1)^{2^{-k}} = n (n+1)^{e_n}
$$
where
$$
e_n = \sum_{k = 1}^{n(n-1)} \frac{1}{2^k} = 1 - 2^{-n^2 + n}.
$$
Thus we need to prove that
$$
\log(n) \leq e_n \log(n+1)
$$
or, in other words
$$
2^{-n^2+n} \log(n+1) \leq \log(n+1) - \log(n)
= \log\left(1 + \frac{1}{n}\right)
$$
and rearranging the terms this becomes
$$
0 \leq \left(2^{n^2-n} - 1\right) \log\left(1 + \frac{1}{n}\right) - \log(n).
$$
The derivative of the rhs is
$$
2^{n^{2} - n} {\left(2 \, n - 1\right)} \log\left(2\right) \log\left(\frac{1}{n} + 1\right) - \frac{2^{n^{2} - n} - 1}{n (n+1)} - \frac{1}{n}
$$
which (squinting a bit) can be seen to be positive for every $n \geq 2$. Finally
$$
\left(2^{2^2-2} - 1\right) \log\left(1 + \frac{1}{2}\right) - \log(2) = 3\log(3) - 4\log(2) = \log\left(\frac{27}{16}\right) > 0
$$
is enough to conclude that $n^2 \leq S(n)^2$ for every $n \geq 2$.
