Problem #2068

 2068 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? $\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 3\qquad \mathrm{(E) \ } 4$ This problem is copyrighted by the American Mathematics Competitions.
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