real analysis - limit of $a_n$ when n to infinity. $a_1=sqrt{k}$ and $a_n = sqrt{k}^{a_{n-1}}$ . $0

I find a question on quora: limit of a sequence.

Generalized Case 1

When you generalize this question like:

\begin{align}
a_1 &= \sqrt{k} \\
a_n &= \sqrt{k}^{a_{n-1}}
\end{align}
where $k =2$, Then: $$\lim_{n \to \infty}a_n =2$$

But what if you change $k$:

1. Well, when $k =3$, it will be $\infty$,(am i wrong about this?). So I am wondering that it will be a number $\alpha$, when $2
   
   
2. And what if $0I do do some research about it. It is bounded, but not monotone. I plot this sequence which $k = \frac{1}{2}$ in mathematica, it seem convergence.

Generalized Case 2

If you treat original quora question as a special case of :
\begin{align}
a_1 &= k^{\frac{1}{k}} \\
a_n &= \left(k^{\frac{1}{k}} \right)^{a_{n-1}}

\end{align}
where $k=2$, Then $\lim_{n \to \infty}a_n =2$ also is true.

Now, consider $k>1$, Then how do I calculate:$$\lim_{n\to \infty}a_n$$

Thanks in advance.