An algebra problem by Nashita Rahman

Let's solve this problem in general, for n > 0 .

$A = \sqrt { n \sqrt { n \sqrt { n \sqrt { n \ldots } } } }$

Then:

$A = \sqrt { nA }$

$A^2 = nA$

This yields us A = 0 (extraneous solution, as it is easy to see, that A > 0)

and A = n (correct solution).

$\text {In our case n=9, hence our answer should be: } \boxed {9}$