I don't know who first observed this (maybe Archimedes?) but it is true because $C({0,1 }^Gamma)$ is a quotient of $ell_1^Gamma$ and hence $ell_1(2^Gamma)$ embeds into $ell_infty(Gamma)$.
@Ady
Here is a more serious answer to your question. Take a quotient map $Q$ from $ell_1(2^Gamma)$ onto $C([0,1]^{2^Gamma})$ and extend to a norm one mapping $T$ from $ell_infty(Gamma)$ into some injective space $Z$ that contains $C([0,1]^{2^Gamma})$ (you cannot extend $Q$ to an operator from $ell_infty(Gamma)$ into $C([0,1]^{2^Gamma})$ because, e.g., $C([0,1]$ is not a quotient of $ell_infty$). Use partitions of unity to get a net $(P_a)$ of norm one finite rank projections on $Z$ taking values in $C([0,1]^{2^Gamma})$ and whose restrictions to $C([0,1]^{2^Gamma})$ converge strongly to the identity. A weak$^*$ cluster point of $(P_a^* T^*)$ gives an isometric embedding of the dual of $C([0,1]^{2^Gamma})$ (which contains $L_1([0,1]^{2^Gamma})$) into the dual of $ell_infty(Gamma)$. Thus if $Y^*$ is any reflexive subspace of $L_1([0,1]^{2^Gamma})$, such as $ell_2(2^Gamma)$, then $Y$ is isometric to a quotient of $ell_infty(Gamma)$.
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