Let $(P,M,\pi,G)$ be a principal bundle and $\omega \in \Omega^{1}(P,\mathfrak{g})$ a principal connection. Given a representation $\rho : G \to GL(V)$ and an equivariant form $\eta \in \Omega^{r}(P,V)$, i.e., $R^{*}_{g} \eta= \rho(g^{-1})\eta$ for all $g \in G$, we have \begin{align*} D \eta = d \eta + \omega \wedge_{\rho'} \eta \end{align*} where $D \eta := (d \eta) h$, $\rho' := \rho_{*e} : \mathfrak{g} \to \mathfrak{gl}(V)$ and \begin{align*} (\omega \wedge_{\rho'} \eta)(X_{1},\dots,X_{r+1}) := \frac{1}{(1+r)!} \sum_{\sigma \in S_{r+1}} \rho'(\omega(X_{\sigma(1)})) \eta(X_{\sigma(2)},\dots,X_{\sigma(r+1)}) \end{align*} for $X_{i} \in T_{u}(P)$, $u \in P$.
Everywhere I look for a proof of this I only find for $\eta = \omega$ or it's put as an exercise. I know how to show for a general rep $\rho$ and $r = 0$, and my guess is that maybe there's some induction to be done, but I have no idea (when opening everything it looks sort of like the functional formula for exterior derivative in terms of brackets idk). Even trying to transport the proof for the case $\eta = \omega$ to general rep with $r=1$ isn't working. Help pls.
I think I got it now. It wasn't induction.
So, for a $0$-form $\nu \in \Omega^{0}(P,V)$ equivariant, take a vector $X \in T_{u}(P)$ and write $vX$ and $hX$ for it's vertical and horizontal component. Then \begin{align} D \nu (X) = d \nu (hX) = d \nu (X-vX) = d \nu(X) - (vX)\nu. \end{align} As $P$ is a principal bundle, there is a unique $\xi \in \mathfrak{g}$ such that $vX = \frac{d}{dt} u \cdot \exp(t \xi) \Big|_{t=0}$. Lets denote by $\xi^{\wedge}$ the fundamental vector field of $\xi$, that is $\xi^{\wedge}_{p}=\frac{d}{dt} p \cdot \exp(t \xi) \Big|_{t=0}$ on $p \in P$. We have \begin{align} (vX)\nu &= (\xi^{\wedge}\nu)_{u}= \frac{d}{dt} \nu(u \cdot \exp(t\xi)) \Big|_{t=0}\\ &= \frac{d}{dt} (R^{*}_{\exp(t\xi)}\nu)_{u} \Big|_{t=0}\\ &= \frac{d}{dt} \rho(\exp(-t\xi))\nu_{u} \Big|_{t=0}\\ &= -\rho'(\xi)\nu_{u}. \end{align} As $\xi = \omega(\xi^{\wedge})= \omega(X)$, the formula follows. For the $r$-form $\eta$, take a coframe $(\epsilon^{1},\dots,\epsilon^{n})$ on $U \subseteq P$. For each $J=(j_{1},\dots,j_{r})$ with $1 \leq j_{1} < \cdots < j_{r} \leq \dim P$ there is a $0$-form $\nu_{J} \in \Omega^{0}(U,V)$ such that \begin{align} \eta|_{U} = \sum_{J} \epsilon^{J} \wedge \nu_{J} \end{align} where $\epsilon^{J}=\epsilon^{j_{1}} \wedge \cdots \wedge \epsilon^{j_{r}}$. It follows that \begin{align} D\eta|_{U} &= \sum_{J} D(\epsilon^{J} \wedge \nu_{J})\\ &= \sum_{J} [ d \epsilon^{J} \wedge \nu_{J} + (-1)^{r} \epsilon^{J} \wedge D\nu_{J}]\\ &= \sum_{J} [ d \epsilon^{J} \wedge \nu_{J} + (-1)^{r} \epsilon^{J} \wedge (d\nu_{J} + \omega \wedge_{\rho'} \nu_{J})]\\ &= \sum_{J} [ d\epsilon^{J} \wedge \nu_{J} + (-1)^{r} \epsilon^{J} \wedge d\nu_{J} + (-1)^{r} \epsilon^{J} \wedge\omega \wedge_{\rho'} \nu_{J}]\\ &= \sum_{J} [ d(\epsilon^{J} \wedge \nu_{J}) + (-1)^{r} \epsilon^{J} \wedge\omega \wedge_{\rho'} \nu_{J}]\\ &= \sum_{J} [ d(\epsilon^{J} \wedge \nu_{J}) + \omega \wedge_{\rho'} (\epsilon^{J} \wedge \nu_{J})]\\ &= [d\eta + \omega \wedge_{\rho'} \eta]|_{U}. \end{align} So we have the equation on the whole $P$, as $(\epsilon^{1},\dots,\epsilon^{n})$ is an arbitrary coframe.