AMC 12 2002a Problem 3

According to the standard convention for exponentiation, $$2^{2^{2^{2}}} = 2^{\left(2^{\left(2^2\right)}\right)} = 2^{16} = 65536.$$

If the order in which the exponentiations are performed is changed, how many other values are possible?

$\mathrm{(A) \ } 0 \mathrm{(B) \ } 1 \mathrm{(C) \ } 2 \mathrm{(D) \ } 3 \mathrm{(E) \ } 4$