Let $E\to M$ be a vector bundle over a smooth (compact) manifold $M$. Let $\nabla^E$ be a connection on $E$. Given a smooth map $f\colon N\to M$, where $N$ is another smooth (compact) manifold, we have the pullback connection $\nabla^{f^*E}$ on the pullback vector bundle $f^*E$.
$\nabla^{f^*E}$ is uniquely determined by
$\nabla^{f^*E}_X(f^*s)=(\nabla^E_{df(X)}s)\circ f$
where $X\in TN$ and $s$ is a section of $E$.
Now, regarding the curvature, do we have
$R^{f^*E}(X,Y)s=R^E(df(X),df(Y))s$
for $X,Y\in TN$, $s$ a section of $f^*E$?
If yes, what is the best way to prove this identity? Can it be shown by just using the definition of curvature ($\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]}$) and a computation in local coordinates? To me that seems messy, but maybe there is some way to simplify the calcuations that I don't see.