Given, $T(n) = \sqrt{n} T(n) + n$.
After the first substitution,we notice that:
$T(n) = \sqrt{n}[n^{1/4} T(n^{1/4}) + \sqrt{n}] + n \implies T(n) = \sqrt{n}[n^{1/4} T(n^{1/4})] + 2n $.
Continuing this series, after $k$ substitutions, we get:
$T(n) = n^{\sum_{i=1}^{k}\frac{1}{2^i}}T(n^{1/2^{k}}) + kn$.
Assuming base case as two, we'll get $k = \log \log n$.
After this, it's pretty much only calculations. Let me know if you don't get it, I'll type it out.