By considering the classifying map $f \colon (\mathbb{C} P^{\infty})^n \rightarrow BU(n)$, its induced map on cohomology, and using the Cartan formula, we can derive the Wu formula for the action of the Steenrod algebra on the cohomology ring $H^*(BU; \mathbb{Z}_2)$, namely, $Sq^i(c_j) = \sum^i_{t=0} {{j+t-i-1} \choose {t} } c_{i-t} c_{j+t}.$ The actions on the cohomology ring of $BU(n)$ come from this, by setting $c_k = 0$ for k > n.
This is just the "BU$ version" of the famous(?) Wu formula for the action of the Steenrod squares on $BO$. But what if we consider the $\mathbb{Z}_p$ cohomology of $BU$, that is, the ring of elementary symmetric polynomials with coefficients in $\mathbb{Z}_p$? How do the Steenrod powers act here? I've had a miserable time trying to use the Cartan formula like we might in the mod $2$ case, and a trip onto google suggests that it's a much more complicated question. The answers I find there are confusing and slightly obfuscated, often coming after pages of mathematics that I'm not really equipped to tackle. I suppose the question is: is there a nice mod $p$ Wu formula our there, in the sense that the mod $2$ formula above is nice (for instance, you can understand it by simply reading it)?