Can we calculate $ i\sqrt { i\sqrt { i\sqrt { \cdots } } }$? It might be obvious that $2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { \cdots } } } } } } $ equals $4.$ So what about $i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } } \text{ ?} $ The answer might be $-1$, but I'm not sure as $i$ is not a real number. Can anyone help?

\begin{eqnarray*}
x= a\sqrt { a\sqrt { a\sqrt { a\sqrt { a\sqrt { a\sqrt { \cdots } } } } } } \\
x=a^{ 1+1/2+1/4+1/8+\cdots} \\
x=a^2
\end{eqnarray*}
So it would seem that
\begin{eqnarray*}
i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } }=\color{red}{-1}.
\end{eqnarray*}
