How do you calculate $ 2^{2^{2^{2^{2}}}} $? From information I have gathered online, this should be equivalent to $2^{16}$ but when I punch the numbers into this large number calculator, the number comes out to be over a thousand digits. Is the calculator wrong or is my method wrong?
 A: This looks an awfully close to what is known as a tetration (a.k.a. power tower).  This is $^{(k)}a=a^{^{(k-1)}a}$ where $^1a=a$. For numbers greater than one, these usually get really big really fast, and faster than exponents do. So in your case, you have $^52=2^{2^{16}}$. Now if you want to see an interesting one look at $\lim_{k\to \infty}\;^{(k)}(\sqrt{2})$.
A: $$2^{2^{2^{2^2}}}=2^{2^{2^4}}=2^{2^{16}}=2^{65536}\tag1$$
The number of digits:
$$\mathcal{A}=1+\lfloor\log_{10}\left(2^{65536}\right)\rfloor=19729\tag2$$
A: Your equation can be simplified using Knuth's up arrow notation.
\begin{equation*}
2^{2^{2^{2^2}}} = 2 \uparrow\uparrow 5
\end{equation*}
(because we can calculate tetration with Knuth's up arrow notation)
By definition of Knuth's up arrow notation, You can get this result.
\begin{equation*}
2\uparrow\uparrow5 = 2^{(2^{(2^{(2^2)})})}
\end{equation*}
And according to web2.0calc,
\begin{equation*}
2^{(2^{(2^{2})})} =  65536
\end{equation*}
Finally, the answer would be:
\begin{equation*}
2^{65536}
\end{equation*}
(correct me if I'm wrong, this was my first answer on Mathematics SE)
A: What you have is a power tower or "tetration" (defined as iterated exponentiation). From the latter link, you would most benefit from this brief excerpt on the difference between iterated powers and iterated exponentials. 
The comment by JMoravitz really gets to the heart of the matter, namely that exponential towers must be evaluated from top to bottom (or right to left). There actually is a notation for your particular question: ${}^52=2^{2^{2^{2^{2}}}}$. You really need to look at ${}^42$ before you get something meaningful because, unfortunately,
$$
{}^32=2^{2^{2}}=2^4=16=4^2=(2^2)^2;
$$
however,
$$
{}^42=2^{2^{2^{2}}}=2^{2^{4}}=2^{16}\neq2^8=(4^2)^2=((2^2)^2)^2.
$$
Hence, your method is wrong, but everything in those links should provide more than enough for you to become comfortable with tetration.
A: By convention, the meaning of things written
$ \displaystyle a^{b^{c^d}} $
without brackets is
$ \displaystyle a^{\left(b^{\left(c^d\right)}\right)} $
and not $\left(\left(a^b\right)^c\right)^d$.
This is because $\left(\left(a^b\right)^c\right)^d$ equals $a^{b\cdot c\cdot d}$ anyway, so it makes pragmatic sense to reserve the raw power-tower notation
$ \displaystyle a^{b^{c^d}} $
for the case that doesn't have an alternative notation without parentheses.
As others have explained, $\displaystyle 2^{2^{2^{2^2}}}$ interpreted with this convention is $2^{65536}$, a horribly huge number, whereas $(((2^2)^2)^2)^2$ is $2^{16}=65536$, as you compute.
