[proofplan]
The proof identifies the derivative at $r = 0$ of the parallel-volume polynomial. The [Steiner formula](/theorems/4122) writes $\mathcal{L}^n(K+rB)$ as a polynomial in $r$, whose linear coefficient is $nW_1(K)$. After subtracting the constant term and dividing by $r$, all terms of degree at least two vanish in the limit $r \downarrow 0$. The equality $W_1(K)=V(K,\dots,K,B)$ is the mixed-volume interpretation of the same coefficient.
[/proofplan]
[step:Expand the parallel volume by the Steiner polynomial]
By the Steiner formula for convex bodies (citing a result not yet in the wiki: Steiner formula for convex bodies), the map $\varphi_K : [0,\infty) \to [0,\infty)$ defined by
\begin{align*}
\varphi_K(r) = \mathcal{L}^n(K+rB)
\end{align*}
has the polynomial expansion
\begin{align*}
\mathcal{L}^n(K+rB)
= \sum_{j=0}^{n} \binom{n}{j} W_j(K) r^j
= W_0(K) + nW_1(K)r + \sum_{j=2}^{n} \binom{n}{j} W_j(K) r^j.
\end{align*}
Evaluating the same identity at $r=0$ gives
\begin{align*}
\mathcal{L}^n(K)=\mathcal{L}^n(K+0B)=W_0(K).
\end{align*}
[guided]
We need to compute the first-order change of the volume of $K+rB$ at $r=0$. The relevant input is the Steiner formula for convex bodies, which says that the parallel volume is a polynomial in the expansion radius $r$. Applied to the convex body $K \subset \mathbb{R}^n$ and the Euclidean unit ball $B$, it gives
\begin{align*}
\mathcal{L}^n(K+rB)
= \sum_{j=0}^{n} \binom{n}{j} W_j(K) r^j.
\end{align*}
We isolate the constant and linear terms because the surface area is defined by a first difference quotient:
\begin{align*}
\mathcal{L}^n(K+rB)
= W_0(K) + nW_1(K)r + \sum_{j=2}^{n} \binom{n}{j} W_j(K) r^j.
\end{align*}
At $r=0$, the parallel body is $K+0B=K$, so the same identity gives
\begin{align*}
\mathcal{L}^n(K)=W_0(K).
\end{align*}
This identifies the term that will cancel in the difference quotient.
[/guided]
[/step]
[step:Take the first variation at $r=0$]
For every $r>0$, subtracting $\mathcal{L}^n(K)=W_0(K)$ from the Steiner expansion and dividing by $r$ gives
\begin{align*}
\frac{\mathcal{L}^n(K+rB)-\mathcal{L}^n(K)}{r}
= nW_1(K) + \sum_{j=2}^{n} \binom{n}{j} W_j(K) r^{j-1}.
\end{align*}
Since each exponent $j-1$ is positive for $j \geq 2$, each term $r^{j-1}$ tends to $0$ as $r \downarrow 0$. Therefore
\begin{align*}
S(K)
= \lim_{r \downarrow 0} \frac{\mathcal{L}^n(K+rB)-\mathcal{L}^n(K)}{r}
= nW_1(K).
\end{align*}
[guided]
The first variation is exactly the right derivative of the parallel-volume function at $0$. Using the expansion from the previous step and the identity $\mathcal{L}^n(K)=W_0(K)$, we compute for $r>0$:
\begin{align*}
\frac{\mathcal{L}^n(K+rB)-\mathcal{L}^n(K)}{r}
&= \frac{W_0(K) + nW_1(K)r + \sum_{j=2}^{n} \binom{n}{j} W_j(K) r^j - W_0(K)}{r} \\
&= nW_1(K) + \sum_{j=2}^{n} \binom{n}{j} W_j(K) r^{j-1}.
\end{align*}
The constant term cancels because it is exactly the volume of $K$. The linear term survives division by $r$ and becomes $nW_1(K)$. Every higher-order term contains a positive power of $r$ after division by $r$, so
\begin{align*}
\lim_{r \downarrow 0} \sum_{j=2}^{n} \binom{n}{j} W_j(K) r^{j-1} = 0.
\end{align*}
Thus the defining first variation satisfies
\begin{align*}
S(K)
= \lim_{r \downarrow 0} \frac{\mathcal{L}^n(K+rB)-\mathcal{L}^n(K)}{r}
= nW_1(K).
\end{align*}
[/guided]
[/step]
[step:Identify the first quermassintegral with the mixed volume coefficient]
By the definition of the mixed-volume coefficients in the Minkowski polynomial expansion, the coefficient $W_1(K)$ in the Steiner formula is
\begin{align*}
W_1(K)=V(K,\dots,K,B),
\end{align*}
where $K$ appears $n-1$ times and $B$ appears once. Substituting this identity into the result of the previous step gives
\begin{align*}
S(K)=nW_1(K)=nV(K,\dots,K,B).
\end{align*}
This is the desired formula.
[/step]
