A youtube video suggests an equation $\left(1+\frac{1}{x}\right)^{x+1}=\left(1+\frac{1}{7}\right)^7$ and solves it in the following way: $$ 1+\frac{1}{x} = \frac{x+1}{x} = \left(\frac{x}{x+1}\right)^{-1}=\left(1-\frac{1}{x+1}\right)^{-1} $$ Therefore: $$ \left(1+\frac{1}{x}\right)^{x+1} = \left(1-\frac{1}{x+1}\right)^{-(x+1)}\ $$ Assuming $t=-(x+1)$ gives: $$ \left(1+\frac{1}{t}\right)^{t}=\left(1+\frac{1}{x}\right)^{x+1}=\left(1+\frac{1}{7}\right)^7 $$ Hence $t=7$ and $x=-8$.
However can we really assume that $t=7$, given that $t$ may be either positive or negative?
Even if we take for granted that function $f(t)=\left(1+\frac{1}{t}\right)^t$ is monotonically increasing (which isn't really obvious), it isn't continuous being undefined for $t\in[-1,0]$ (where $1+\frac1t$ is undefined or non-positive), therefore I don't see a reason to claim that the solution is unique.