What is $\Gamma(1+\Gamma(1+\Gamma(1+\dots)))$? Consider the infinitely nested expression
$$x=\Gamma(1+\Gamma(1+\Gamma(1+\dots)))$$
where $\Gamma$ is the Gamma function.
Imitating the standard method for solving infinitely nested radicals, we can write
$$\Gamma(1+x)=x$$
and solve for $x$. This yields two positive solutions: $1$ and $2$.
If we instead imagine that the "$\dots$" part of the "innermost term" (hand waving here) disappears then that term becomes $\Gamma(1)=1$, the surrounding term becomes $\Gamma(2)=1$ and the hierarchy collapses until $\Gamma(2)=1$ remains.
What, then, about the "solution" $x=2$ we derived using the first method? Is either of these solutions valid? Or is it impossible to assign a unique meaningful value to the infinite expression in the first place?
 A: The infinitely nested gamma functions equation can be solved (as you suggest) by the standard method for infinite nesting:
$$
\begin{align}
x    &= \Gamma(1+\Gamma(1+\Gamma(1+\cdots)))\\
     &= \Gamma(1+x)\\
     &= x\Gamma(x)\\
\Gamma(x)
     &= 1\\
x    &= 1,2
\end{align}
$$
Plugging in $x=1$ solves the equation:
$$
\begin{align}
x    &= \Gamma(1+1)\\\
     &= \Gamma(2)\\
     &= 1
\end{align}
$$
And plugging in $x=2$ also solves the equation:
$$
\begin{align}
x    &= \Gamma(1+2)\\
     &= \Gamma(3)\\
     &= 2\Gamma(2)\\
     &= 2
\end{align}
$$
So both $x=1$ and $x=2$, are the solutions for the original equation.

Your hand-waving argument is invalid, because you are effectively taking the assumption that $(\cdots=)\ x=0,$ and then proving that $x=1$:
$$
\color{grey}{
\begin{align}
x    &= \Gamma(1+\Gamma(1+\Gamma(1+\cdots)))\\
     &= \Gamma(1+\Gamma(1+\Gamma(1+0)))\\
     &= \Gamma(1+\Gamma(1+\Gamma(1)))\\
     &= \Gamma(1+\Gamma(1+1))\\
     &= \Gamma(1+\Gamma(2))\\
     &= \Gamma(2)\\
     &= 1
\end{align}
}
$$

A similar hand-waving argument for the $x=1$ case is:
$$
\begin{align}
x    &= \Gamma(1+\Gamma(1+\Gamma(1+\cdots)))\\
     &= \Gamma(1+\Gamma(1+\Gamma(1+1)))\\
     &= \Gamma(1+\Gamma(1+\Gamma(2)))\\
     &= \Gamma(1+\Gamma(1+1))\\
     &= \Gamma(1+\Gamma(2))\\
     &= \Gamma(1+1)\\
     &= \Gamma(2)\\
     &= 1
\end{align}
$$
And a hand-waving argument for the $x=2$ case is:
$$
\begin{align}
x    &= \Gamma(1+\Gamma(1+\Gamma(1+\cdots)))\\
     &= \Gamma(1+\Gamma(1+\Gamma(1+2)))\\
     &= \Gamma(1+\Gamma(1+\Gamma(3)))\\
     &= \Gamma(1+\Gamma(1+2))\\
     &= \Gamma(1+\Gamma(3))\\
     &= \Gamma(1+2)\\
     &= \Gamma(3)\\
     &= 2
\end{align}
$$
Unfortunately, both of these hand-waving arguments are indeed circular.
