The Kudo Trangression Theorem has to do with the transgression in the Leray-Serre spectral sequence for cohomology in $\mathbb{Z}/p$ ($p$ odd). It can be proved by the method of the universal example, once it is shown that in the path-loop fibration sequence $K(\mathbb{Z}/p,2n) \to P(K(\mathbb{Z}/p,2n+1)) \to K(\mathbb{Z}/p,2n+1)$

the fundamental class $v$ of the fiber transgresses to $u$, that of the base

this forces a zig-zag of cancellation, up to $v^{p-1}\mapsto u \otimes v^{p-2}$

also $v^p$ transgresses to $P^n(u)$, and

$u\otimes v^{p-1}$ "transgresses" to $\beta P^n(u)$.

Parts (1), (2) and (3) are easy, but part (4) seems difficult. There is a proof along these lines in a paper of Browder from the mid 1960's (he attributes the proof to Milgram), but the proof of (4) is actually quite hard and leans heavily on algebraic mucking around in the spectral sequence.

Does anyone know of a clever way to prove (4)?

Edit: Let's say we know by induction that the cohomology of the fiber is what it has to be. Then I think the behavior of the spectral sequence is forced in dimensions below that of $u\otimes v^{p-1}$. Does this show that $u\otimes v^{p-1}$ "transgresses"? Suppose it does; then its image is $Q(u)$, where $Q$ is a cohomology operation that vanishes when looped (since it is not the transgression of a class in the fiber). Perhaps we can argue that $Q$ must be $\beta P^n$, up to sign?