[proofplan]
The tautological bundle $S$ sits as a subbundle of the product bundle $\underline{\mathbb C}^n$ over the Grassmannian, and the quotient is precisely $Q$. Applying the Whitney product formula to this short exact sequence gives $c(S)c(Q)=c(\underline{\mathbb C}^n)$. The product bundle has total Chern class $1$, so the desired multiplicative relation follows. Expanding the product degree by degree gives the equivalent relations among the individual Chern classes.
[/proofplan]
[step:Identify the tautological short exact sequence]
Let $X:=\operatorname{Gr}(k,n)$, let $\pi_S:S\to X$ denote the tautological bundle, let $\pi_{\mathrm{triv}}:\underline{\mathbb C}^n=X\times\mathbb C^n\to X$ denote the product bundle, and let $\pi_Q:Q\to X$ denote the quotient bundle. For each point $V\in X$, the inclusion of the fiber $S_V=V$ into $\mathbb C^n$ gives a fiberwise injective bundle map
\begin{align*}
\iota:S\longrightarrow \underline{\mathbb C}^n.
\end{align*}
By definition of the quotient bundle, $Q$ is the cokernel bundle of this inclusion, with fiber
\begin{align*}
Q_V=\mathbb C^n/V.
\end{align*}
Hence there is a short exact sequence of complex vector bundles over $X$:
\begin{align*}
0\longrightarrow S\stackrel{\iota}{\longrightarrow}\underline{\mathbb C}^n\longrightarrow Q\longrightarrow 0.
\end{align*}
[/step]
[step:Apply the Whitney product formula to the tautological exact sequence]
We use the [Whitney product formula for Chern classes](/theorems/7052) [citetheorem:9797]: if
\begin{align*}
0\longrightarrow E'\longrightarrow E\longrightarrow E''\longrightarrow 0
\end{align*}
is a short exact sequence of complex vector bundles over a space $Y$, then
\begin{align*}
c(E)=c(E')c(E'')
\end{align*}
in $H^*(Y;\mathbb Z)$. The sequence from the previous step is a short exact sequence of complex vector bundles over $X$, so the formula applies with $E'=S$, $E=\underline{\mathbb C}^n$, and $E''=Q$. Therefore
\begin{align*}
c(\underline{\mathbb C}^n)=c(S)c(Q).
\end{align*}
[guided]
The purpose of this step is to convert the geometric exact sequence into an identity in cohomology. The relevant input is the Whitney product formula for Chern classes [citetheorem:9797], which says that for every short exact sequence of complex vector bundles
\begin{align*}
0\longrightarrow E'\longrightarrow E\longrightarrow E''\longrightarrow 0
\end{align*}
over a space $Y$, the total Chern classes satisfy
\begin{align*}
c(E)=c(E')c(E'')
\end{align*}
in the [cohomology ring](/theorems/2271) $H^*(Y;\mathbb Z)$.
We verify the hypotheses in the present situation. The base space is $Y=X=\operatorname{Gr}(k,n)$. The bundle $S\to X$ is a complex vector bundle because its fiber over $V\in X$ is the complex [vector space](/page/Vector%20Space) $V$. The bundle $\underline{\mathbb C}^n=X\times\mathbb C^n\to X$ is a complex vector bundle. The bundle $Q\to X$ is the quotient complex vector bundle with fiber $Q_V=\mathbb C^n/V$. The maps
\begin{align*}
0\longrightarrow S\stackrel{\iota}{\longrightarrow}\underline{\mathbb C}^n\longrightarrow Q\longrightarrow 0
\end{align*}
are complex-linear on each fiber, and the sequence is exact on each fiber because $S_V=V\subset \mathbb C^n$ and $Q_V=\mathbb C^n/V$.
Thus the Whitney product formula applies with $E'=S$, $E=\underline{\mathbb C}^n$, and $E''=Q$. It gives
\begin{align*}
c(\underline{\mathbb C}^n)=c(S)c(Q)
\end{align*}
in $H^*(X;\mathbb Z)$.
[/guided]
[/step]
[step:Use the normalization of Chern classes on the product bundle]
The product bundle $\underline{\mathbb C}^n$ has total Chern class
\begin{align*}
c(\underline{\mathbb C}^n)=1.
\end{align*}
This is the normalization property of Chern classes for product complex vector bundles, equivalently the vanishing of all positive-degree Chern classes of such a bundle. Substituting this into the identity obtained above gives
\begin{align*}
c(S)c(Q)=1
\end{align*}
in $H^*(X;\mathbb Z)$.
[/step]
[step:Read off the positive-degree relations]
Write
\begin{align*}
c(S)=1+s_1+\cdots+s_k
\end{align*}
with $s_i:=c_i(S)\in H^{2i}(X;\mathbb Z)$, and write
\begin{align*}
c(Q)=1+q_1+\cdots+q_{n-k}
\end{align*}
with $q_j:=c_j(Q)\in H^{2j}(X;\mathbb Z)$. Define $s_0:=1$, $q_0:=1$, set $s_i:=0$ for $i>k$, and set $q_j:=0$ for $j>n-k$. The degree-$2d$ component of the product $c(S)c(Q)$ is then
\begin{align*}
\sum_{i+j=d}s_iq_j.
\end{align*}
Since $c(S)c(Q)=1$, all components of positive cohomological degree vanish. Therefore, for every integer $d\ge 1$,
\begin{align*}
\sum_{i+j=d}s_iq_j=0
\end{align*}
in $H^{2d}(X;\mathbb Z)$. This is exactly the stated equivalent form of the Chern relation.
[/step]
