In [1] Berstein and Ganea define the nilpotence of an H-space to be the least integer n such that the n-commutator is nullhomotopic. We prove that S3 with the usual multiplication is 4 nilpotent. Let X be an H-space. The 2-commutator c2: XXX->X is defined by c2(x, y) = xyx-ly-1 where the multiplication and inverses are given by the H-space structure of X. The n-commutator C.: Xn-?X is defined inductively by Cn=C2(Cn-lX 1). Let T1(X) denote the subset of Xn consisting of those n-tuples (x1, , xn) such that xi= * (the base point) for at least one i. It is well known that cnj T1(X)*. Thus a map c/n X /Tj(X) -X may be defined such that the homotopy class of On depends only upon the homotopy class of cn. Let 4)n be the homotopy class of On. 4)n is the Samelson product and Cn is nullhomotopic if and only if ;tn = 0 The usual multiplication for S3 is that obtained by considering S3 to be the set of unit quaternions. With this multiplication Q., infinite quaternionic projective space, is a classifying space for S3. Let T: 7r(ISn-1, X)-7r(Sn-1, OX) be defined by (Tf(s))(t) =f(t, s), where sCEzSn-1, fE(2Sn-1, X), and tEI. T is an isomorphism of the homotopy groups. Samelson [3] has shown that if j: S4--Q. is inclusion, Tj: S3 -'QQ. is an H-homomorphism which is also a homotopy equivalence. He uses this to show that T[ [j, j], j] = (Tj)*(3 where the product on the left is the 3-fold iterated Whitehead product. Since T is an isomorphism, to show that S3 is 4 nilpotent it suffices to show that the four-fold iterated Whitehead product of j is zero and the threefold product is nonzero. Let i4 be the identity map on S4. Hilton [2] has shown that O 0 [ [i4, i4], i4] EZr1o(S4) is the image of an element in 1rg (S3) under the suspension homomorphism. He uses this fact to prove