Here's another proof which uses mixed exterior algebra. The advantage is that it doesn't involve messy computations with bases, sign factors, determinants, etc. but the disadvantage is that it uses a bunch of other machinery. I summarize facts that are needed below, but for details see Chapter 6 of [1].
Background
As above let $V$ be $n$-dimensional, let $V^*$ be dual to $V$, and define
$$\textstyle\bigwedge(V^*,V)=\bigwedge V^*\otimes\bigwedge V$$
to be the mixed exterior algebra over $V^*,V$, which is the canonical tensor product of the exterior algebras $\bigwedge V^*$ and $\bigwedge V$. The mixed exterior product (or "dot" product) satisfies
$$(u^*\otimes u)\cdot(v^*\otimes v)=(u^*\wedge v^*)\otimes(u\wedge v)$$
for all $u^*,v^*\in\bigwedge V^*$ and $u,v\in\bigwedge V$. Importantly, this product is commutative in the diagonal subalgebra
$$\Delta(V^*,V)=\bigoplus_{p=0}^n\Delta_p(V^*,V)\qquad\text{where}\qquad\Delta_p(V^*,V)=\textstyle\bigwedge^p V^*\otimes\bigwedge^p V$$
to which we restrict attention. For $z\in\Delta(V^*,V)$, we write
$$z^k=\frac{1}{k!}\underbrace{z\cdots z}_{k\text{ factors}}$$
There is an inner product in $\Delta(V^*,V)$ induced by the scalar product between $V^*$ and $V$ which satisfies
$$\langle u^*\otimes u,v^*\otimes v\rangle=\langle u^*,v\rangle\langle v^*,u\rangle$$
There is a canonical linear isomorphism $T:\Delta(V^*,V)\to H(\bigwedge V;\bigwedge V)$ to the space of homogeneous linear transformations of $\bigwedge V$ which satisfies
$$T(u^*\otimes u)(v)=\langle u^*,v\rangle u$$
Making appropriate identifications under $T$, we write
$$\textstyle H(\bigwedge V;\bigwedge V)=\displaystyle\bigoplus_{p=0}^n L(\textstyle\bigwedge^p V;\bigwedge^p V)$$
so in particular all linear transformations of $V$ and their exterior powers are in there. We make $T$ an algebra isomorphism by defining the (generalized) box product
$$\alpha\mathbin{\square}\beta=T(T^{-1}\alpha\cdot T^{-1}\beta)$$
for homogeneous $\alpha,\beta:\bigwedge V\to\bigwedge V$. For $z\in V^*\otimes V$ and $\varphi=T(z):V\to V$, it can be shown that
$$T(z^k)=\frac{1}{k!}T(\underbrace{z\cdots z}_{k\text{ factors}})=\frac{1}{k!}\underbrace{\varphi\mathbin{\square}\cdots\mathbin{\square}\varphi}_{k\text{ factors}}=\textstyle\bigwedge^k\varphi$$
$T$ is also an isometry from the inner product above to the trace form $\langle\alpha,\beta\rangle=\mathop{\mathrm{tr}}(\alpha\circ\beta)$. If we define the unit tensor $t=T^{-1}(\iota)$ where $\iota:V\to V$ is the identity map, then for $z\in\Delta_k(V^*,V)$ it follows that
$$\langle t^k,z\rangle=\mathop{\mathrm{tr}}(T(z))$$
Finally, it can be shown that
$$i(t^k)t^n=t^{n-k}$$
where for $z\in\Delta(V^*,V)$, $i(z)$ is the insertion of $z$, dual to exterior multiplication by $z$:
$$\langle i(z)x,y\rangle=\langle x,z\cdot y\rangle=\langle x,y\cdot z\rangle$$
Solution
For $\varphi:V\to V$ with $z=T^{-1}(\varphi)$, we have
$$\begin{align*}
\textstyle\mathop{\mathrm{tr}}(\bigwedge^k\varphi)&=\langle t^k,z^k\rangle\\
&=\langle i(t^{n-k})t^n,z^k\rangle\\
&=\langle t^n,z^k\cdot t^{n-k}\rangle\\
&=\textstyle\mathop{\mathrm{tr}}(\bigwedge^k\varphi\mathbin{\square}\bigwedge^{n-k}\iota)
\end{align*}$$
Let $\Phi=\bigwedge^k\varphi\mathbin{\square}\bigwedge^{n-k}\iota$. Then $\Phi:\bigwedge^n V\to\bigwedge^n V$, and analysis of the box product shows that
$$\Phi(v_1\wedge\cdots\wedge v_n)=\frac{1}{k!\,(n-k)!}\sum_{\sigma\in S_n}\varphi_{\sigma(1)}v_1\wedge\cdots\wedge\varphi_{\sigma(n)}v_n\tag{1}$$
where
$$\varphi_i=\begin{cases}
\varphi&\text{if }1\le i\le k\\
\iota&\text{if }k<i\le n
\end{cases}$$
A moment's thought shows that (1) is your desired sum, since for every $1\le i_1<\cdots<i_k\le n$, there will be precisely $k!\,(n-k)!$ terms of form
$$v_1\wedge\cdots\wedge\varphi v_{i_1}\wedge\cdots\wedge\varphi v_{i_k}\wedge\cdots\wedge v_n$$
and every term is of one of these forms.
References
- Greub, W. Multilinear Algebra, 2nd ed. Springer, 1978.
