$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.

$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$


Let $$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=x $$
Clearly, $x>0$
$$\implies x^2=7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=7x$$
Now left is the proof of converge(as conversed with  Abdulh Khazzak Gustav ElFakiri)
Observe that the $r$th term $T_r$ of this infinite product is $\displaystyle7^{\left(\frac1{2^r}\right)}$
using Convergence/Divergence of infinite product, $$\sum_{0\le r<\infty}\ln(T_r)=\ln 7\sum_{0\le r<\infty}\frac1{2^r}$$ which is an infinite Geometric Series with common ratio $=\frac12$ which $\in(-1,1)$, hence  the later Series is convergent $\left(\text{ in fact }\displaystyle=\ln7\cdot\frac1{1-\frac12}\right)$, so will be the original infinite Product
