What's $-2^{-2^{-2^{-2...}}}$? Pretty simple question that I know someone has probably asked before: 
What's $-2^{-2^{-2^{-2...}}}$? Or, more specifically, what is the number that this repeated sequence approaches?  I have calculated it out to about $-0.641185$, but I have no idea what that number is or where you could get it from.
Keep in mind that I am not asking for $(-2)^{(-2)^{(-2)...}}$, this sequence doesn't really work. 
The order of operations looks like this: $-(2^{-(2^{-(2^{-(2...)})})})$ 
It's basically like taking $-2$, exponentiating it by 2 to get $2^{-2}$, then making that negative $-2^{-2}$, then exponentiating it again, etc.
 A: Recursively, we might define this as the limit as $n \to \infty$ of
$$a_n = -(2)^{a_{n-1}}$$
with $a_1 = -2$. Letting $n \to \infty$ on the assumptions of continuity and convergence of $a_n$ to some value $L$ gives us
$$L = -2^L$$
This value $L$ is fundamentally what you seek. This equation will have to be solved by means of the Lambert W function and doesn't have a closed form outside of that, to my knowledge anyhow. Note that $-2^L = -e^{\ln(2) \cdot L}$. Thus, follow the following steps:


*

*Use the identity above

*Multiply both sides by $-\ln(2)$

*Multiply both sides by $e^{-\ln(2)L}$
Then you get that
$$L = -2^L \implies -\ln(2) L  e^{-\ln(2) L}= \ln(2)$$
The Lambert W function is the inverse to $f(x) = xe^x$. That is, $W(xe^x) = x$. Applying it to both sides of the above, we obtain that
$$W(-\ln(2) L  e^{-\ln(2) L})= W(\ln(2)) \implies -\ln(2)L = W(\ln(2))$$
Solving for $L$ gives us
$$ L = \frac{-W(\ln(2))}{\ln(2)}$$
Of course, this is about as much simplification as we can get; many values for the W function have to be approximated. Of course, we also should note this is the principle value since it is a multivalued function otherwise. If we approximate the value by, say, Wolfram, we see that
$$L \approx -0.641186$$
