According to the standard convention for exponentiation, 
\[2^{2^{2^{2}}} = 2^{(2^{(2^2)})} = 2^{16} = 65536.\]
If the order in which the exponentiations are performed is changed, how many other values are possible?
$\textbf{(A) } 0\qquad \textbf{(B) } 1\qquad \textbf{(C) } 2\qquad \textbf{(D) } 3\qquad \textbf{(E) } 4$
The best way to solve this problem is by simple brute force. 
It is convenient to drop the usual way how exponentiation is denoted, and to write the formula as $2\uparrow 2\uparrow 2\uparrow 2$, where $\uparrow$ denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so:

$2\uparrow (2\uparrow (2\uparrow 2))$
$2\uparrow ((2\uparrow 2)\uparrow 2)$
$((2\uparrow 2)\uparrow 2)\uparrow 2$
$(2\uparrow (2\uparrow 2))\uparrow 2$
$(2\uparrow 2)\uparrow (2\uparrow 2)$
We can note that $2\uparrow (2\uparrow 2) = (2\uparrow 2)\uparrow 2 =16$. Therefore options 1 and 2 are equal, and options 3 and 4 are equal.
Option 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.
$((2\uparrow 2)\uparrow 2)\uparrow 2 = 16\uparrow 2 = 256$
$(2\uparrow 2)\uparrow (2\uparrow 2) = 4 \uparrow 4 = 256$
Thus the only other result is $256$, and our answer is $\boxed{\textbf{(B) } 1}$.