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Yes. Let $Sq^j$ be a square operation in Steenrod algebra. We have $$Sq^j(x)=\left\{\begin{matrix} x, \ \ \ \text{if $j=0$,}\\ x^2, \ \ \ \text{if $j=1$,}\\ 0. \ \ \ \text{otherwise.} \end{matrix}\right.$$ And it satisfies the Cartan fomular $$Sq^{n}(fg)=\sum_{i=0}^nSq^i(f)Sq^{n-i}(g).$$