According to the standard convention for exponentiation,
2^{2^{2^2}} \equal{} 2^{\left(2^{\left(2^2\right)}\right)} \equal{} 2^{16} \equal{} 65,\!536. If the order in which the exponentiations are performed is changed, how many other values are possible?<spanclass=′latex−bold′>(A)</span>0<spanclass=′latex−bold′>(B)</span>1<spanclass=′latex−bold′>(C)</span>2<spanclass=′latex−bold′>(D)</span>3<spanclass=′latex−bold′>(E)</span>4