$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=7^\frac{1}{2}\cdot7^\frac{1}{4}\cdot 7^\frac{1}{8}\cdots=7^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots}=7^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}=7$$
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
Your expression can be written as $$7^{\frac12 + \frac14 ...}.$$
Now you can use sum of infinite GP = $\frac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio.
Thus sum $= 1$.
Your expression $=$ $7^1$ = $7$