[proofplan] We begin from the antisymmetrisation definition of the wedge product, in which $(\alpha \wedge \beta)(v_1, \dots, v_{p+q})$ is a normalised signed sum over the full symmetric group $S_{p+q}$. The key combinatorial input is the bijection $S_{p+q} \leftrightarrow \operatorname{Sh}(p,q) \times S_p \times S_q$ that decomposes a permutation uniquely as a shuffle followed by internal permutations of the two blocks. The sign is multiplicative under this decomposition. Because $\alpha$ and $\beta$ are alternating, each internal permutation contributes a sign factor that exactly cancels the corresponding sign in the decomposition. Each fixed shuffle therefore appears with multiplicity $p!\,q!$, which cancels the combinatorial prefactor and leaves precisely the claimed shuffle sum. [/proofplan] [step:Recall the antisymmetrisation definition of the wedge product] For an arbitrary multilinear map $T : V^{p+q} \to \mathbb{R}$, define the alternation \begin{align*} \operatorname{Alt}(T) &: V^{p+q} \to \mathbb{R} \\ (v_1, \dots, v_{p+q}) &\mapsto \frac{1}{(p+q)!} \sum_{\sigma \in S_{p+q}} \operatorname{sgn}(\sigma)\, T(v_{\sigma(1)}, \dots, v_{\sigma(p+q)}). \end{align*} The wedge product of $\alpha \in \Lambda^p(V^*)$ and $\beta \in \Lambda^q(V^*)$ is defined by \begin{align*} \alpha \wedge \beta := \frac{(p+q)!}{p!\,q!}\, \operatorname{Alt}(\alpha \otimes \beta), \end{align*} where $(\alpha \otimes \beta)(v_1, \dots, v_{p+q}) = \alpha(v_1, \dots, v_p)\, \beta(v_{p+1}, \dots, v_{p+q})$. Combining the two formulas, we obtain \begin{align*} (\alpha \wedge \beta)(v_1, \dots, v_{p+q}) = \frac{1}{p!\,q!} \sum_{\sigma \in S_{p+q}} \operatorname{sgn}(\sigma)\, \alpha(v_{\sigma(1)}, \dots, v_{\sigma(p)})\, \beta(v_{\sigma(p+1)}, \dots, v_{\sigma(p+q)}). \end{align*} This identity is the starting point of the proof; the remaining work is to reduce the $(p+q)!$-term sum to a sum over $|\operatorname{Sh}(p,q)| = \binom{p+q}{p}$ terms. [/step] [step:Establish the bijection $S_{p+q} \cong \operatorname{Sh}(p,q) \times S_p \times S_q$] Define a map \begin{align*} \Phi: \operatorname{Sh}(p,q) \times S_p \times S_q &\to S_{p+q} \\ (\tau, \pi, \rho) &\mapsto \tau \circ (\pi \oplus \rho), \end{align*} where $\pi \oplus \rho \in S_{p+q}$ is the block-diagonal permutation defined by \begin{align*} (\pi \oplus \rho)(i) = \begin{cases} \pi(i) & \text{if } 1 \le i \le p, \\ p + \rho(i - p) & \text{if } p+1 \le i \le p+q. \end{cases} \end{align*} We claim $\Phi$ is a bijection. Given $\sigma \in S_{p+q}$, let $A := \sigma(\{1, \dots, p\}) \subseteq \{1, \dots, p+q\}$, which has cardinality $p$. There is a unique $\tau \in \operatorname{Sh}(p,q)$ with $\tau(\{1, \dots, p\}) = A$: namely, $\tau(1) < \dots < \tau(p)$ is the increasing enumeration of $A$, and $\tau(p+1) < \dots < \tau(p+q)$ is the increasing enumeration of $\{1, \dots, p+q\} \setminus A$. The composition $\tau^{-1} \circ \sigma$ then maps $\{1, \dots, p\}$ to itself and $\{p+1, \dots, p+q\}$ to itself, so it has the form $\pi \oplus \rho$ for unique $\pi \in S_p$ and $\rho \in S_q$. Hence $\sigma = \tau \circ (\pi \oplus \rho) = \Phi(\tau, \pi, \rho)$, and the decomposition is unique. This proves $\Phi$ is a bijection. [guided] The idea behind the bijection is that every permutation of $\{1, \dots, p+q\}$ is determined by three pieces of data: (i) which $p$-element subset of $\{1, \dots, p+q\}$ becomes the image of the first block $\{1, \dots, p\}$; (ii) how the first block is permuted internally before being mapped to that subset; (iii) how the second block $\{p+1, \dots, p+q\}$ is permuted internally before being mapped to the complement. The shuffle $\tau$ records (i): it sends $\{1, \dots, p\}$ to $A$ via the order-preserving bijection, and sends $\{p+1, \dots, p+q\}$ to $A^c$ via the order-preserving bijection. The conditions $\tau(1) < \dots < \tau(p)$ and $\tau(p+1) < \dots < \tau(p+q)$ are exactly the conditions that pin down a unique such $\tau$ for each choice of $A$. Therefore $|\operatorname{Sh}(p,q)| = \binom{p+q}{p}$. Existence of the decomposition: define $\tau$ as the unique shuffle with $\tau(\{1, \dots, p\}) = \sigma(\{1, \dots, p\})$. Then $\tau^{-1}\sigma$ preserves the partition $\{1, \dots, p\} \sqcup \{p+1, \dots, p+q\}$ setwise, so it acts independently on the two blocks: this is the data of $(\pi, \rho) \in S_p \times S_q$. Uniqueness: if $\tau \circ (\pi \oplus \rho) = \tau' \circ (\pi' \oplus \rho')$, then applying both sides to $\{1, \dots, p\}$ gives $\tau(\{1, \dots, p\}) = \tau'(\{1, \dots, p\})$, so $\tau = \tau'$ (since each shuffle is determined by its image on the first block). Cancelling $\tau$ yields $\pi \oplus \rho = \pi' \oplus \rho'$, hence $\pi = \pi'$ and $\rho = \rho'$. [/guided] [/step] [step:Compute the sign and the function values under the decomposition] Fix $\sigma \in S_{p+q}$ and write $\sigma = \tau \circ (\pi \oplus \rho)$ with $\tau \in \operatorname{Sh}(p,q)$, $\pi \in S_p$, $\rho \in S_q$. Since $\operatorname{sgn}: S_{p+q} \to \{\pm 1\}$ is a group homomorphism, and the block-diagonal $\pi \oplus \rho$ has sign $\operatorname{sgn}(\pi) \operatorname{sgn}(\rho)$, we obtain \begin{align*} \operatorname{sgn}(\sigma) = \operatorname{sgn}(\tau)\, \operatorname{sgn}(\pi \oplus \rho) = \operatorname{sgn}(\tau)\, \operatorname{sgn}(\pi)\, \operatorname{sgn}(\rho). \end{align*} For the function values, observe that $\sigma(i) = \tau(\pi(i))$ for $1 \le i \le p$ and $\sigma(p+j) = \tau(p + \rho(j))$ for $1 \le j \le q$. Hence \begin{align*} \alpha(v_{\sigma(1)}, \dots, v_{\sigma(p)}) = \alpha(v_{\tau(\pi(1))}, \dots, v_{\tau(\pi(p))}). \end{align*} Because $\alpha \in \Lambda^p(V^*)$ is alternating, permuting its arguments by $\pi \in S_p$ produces a factor of $\operatorname{sgn}(\pi)$: \begin{align*} \alpha(v_{\tau(\pi(1))}, \dots, v_{\tau(\pi(p))}) = \operatorname{sgn}(\pi)\, \alpha(v_{\tau(1)}, \dots, v_{\tau(p)}). \end{align*} Similarly, since $\beta \in \Lambda^q(V^*)$ is alternating, \begin{align*} \beta(v_{\sigma(p+1)}, \dots, v_{\sigma(p+q)}) = \beta(v_{\tau(p+\rho(1))}, \dots, v_{\tau(p+\rho(q))}) = \operatorname{sgn}(\rho)\, \beta(v_{\tau(p+1)}, \dots, v_{\tau(p+q)}). \end{align*} Multiplying these three identities yields \begin{align*} &\operatorname{sgn}(\sigma)\, \alpha(v_{\sigma(1)}, \dots, v_{\sigma(p)})\, \beta(v_{\sigma(p+1)}, \dots, v_{\sigma(p+q)}) \\ &\qquad = \operatorname{sgn}(\tau)\, \operatorname{sgn}(\pi)\, \operatorname{sgn}(\rho)\, \cdot\, \operatorname{sgn}(\pi)\, \alpha(v_{\tau(1)}, \dots, v_{\tau(p)})\, \cdot\, \operatorname{sgn}(\rho)\, \beta(v_{\tau(p+1)}, \dots, v_{\tau(p+q)}) \\ &\qquad = \operatorname{sgn}(\tau)\, \alpha(v_{\tau(1)}, \dots, v_{\tau(p)})\, \beta(v_{\tau(p+1)}, \dots, v_{\tau(p+q)}), \end{align*} where the last equality uses $\operatorname{sgn}(\pi)^2 = \operatorname{sgn}(\rho)^2 = 1$. [guided] This is the heart of the argument: the internal-permutation signs cancel. When we decompose $\sigma = \tau \circ (\pi \oplus \rho)$, the sign of $\sigma$ acquires three factors: $\operatorname{sgn}(\tau)$ from the shuffle, $\operatorname{sgn}(\pi)$ from the internal permutation of the first block, and $\operatorname{sgn}(\rho)$ from the internal permutation of the second block. Each of $\operatorname{sgn}(\pi)$ and $\operatorname{sgn}(\rho)$ looks like a problem: they make the summand depend on $(\pi, \rho)$, not just on the shuffle $\tau$. But $\alpha$ is alternating, so reshuffling its arguments by $\pi$ produces an identical sign factor $\operatorname{sgn}(\pi)$. The signs from the decomposition of $\sigma$ and from the alternating property of $\alpha$ are the *same* sign, so they multiply to $\operatorname{sgn}(\pi)^2 = 1$ and disappear. Likewise for $\beta$ and $\rho$. What survives is exactly $\operatorname{sgn}(\tau)\, \alpha(v_{\tau(1)}, \dots, v_{\tau(p)})\, \beta(v_{\tau(p+1)}, \dots, v_{\tau(p+q)})$ — a quantity that depends only on the shuffle $\tau$, not on the internal permutations. This is why the sum collapses from $|S_{p+q}| = (p+q)!$ terms to $|\operatorname{Sh}(p,q)| = \binom{p+q}{p}$ distinct values, each appearing $p!\,q!$ times. Note the role of the alternating hypothesis: without it, we would still have the decomposition of $\operatorname{sgn}(\sigma)$, but the function-value rearrangements would *not* produce compensating signs, and the formula would fail. The shuffle formula is genuinely a consequence of alternation. [/guided] [/step] [step:Reindex the sum over $S_{p+q}$ as a sum over shuffles with multiplicity $p!\,q!$] Using the bijection $\Phi$ from the previous step, we reindex the sum over $S_{p+q}$: \begin{align*} &\sum_{\sigma \in S_{p+q}} \operatorname{sgn}(\sigma)\, \alpha(v_{\sigma(1)}, \dots, v_{\sigma(p)})\, \beta(v_{\sigma(p+1)}, \dots, v_{\sigma(p+q)}) \\ &\qquad = \sum_{\tau \in \operatorname{Sh}(p,q)} \sum_{\pi \in S_p} \sum_{\rho \in S_q} \operatorname{sgn}(\tau)\, \alpha(v_{\tau(1)}, \dots, v_{\tau(p)})\, \beta(v_{\tau(p+1)}, \dots, v_{\tau(p+q)}), \end{align*} where we used the computation from the previous step to evaluate each summand. The inner summand no longer depends on $(\pi, \rho)$, so the inner double sum contributes a multiplicative factor of $|S_p| \cdot |S_q| = p!\, q!$: \begin{align*} \sum_{\sigma \in S_{p+q}} \operatorname{sgn}(\sigma)\, \alpha(v_{\sigma(1)}, \dots, v_{\sigma(p)})\, \beta(v_{\sigma(p+1)}, \dots, v_{\sigma(p+q)}) = p!\, q! \sum_{\tau \in \operatorname{Sh}(p,q)} \operatorname{sgn}(\tau)\, \alpha(v_{\tau(1)}, \dots, v_{\tau(p)})\, \beta(v_{\tau(p+1)}, \dots, v_{\tau(p+q)}). \end{align*} [/step] [step:Combine to recover the shuffle formula] Substituting the previous identity into the expression for $(\alpha \wedge \beta)(v_1, \dots, v_{p+q})$ from the first step, we obtain \begin{align*} (\alpha \wedge \beta)(v_1, \dots, v_{p+q}) &= \frac{1}{p!\,q!} \cdot p!\, q! \sum_{\tau \in \operatorname{Sh}(p,q)} \operatorname{sgn}(\tau)\, \alpha(v_{\tau(1)}, \dots, v_{\tau(p)})\, \beta(v_{\tau(p+1)}, \dots, v_{\tau(p+q)}) \\ &= \sum_{\tau \in \operatorname{Sh}(p,q)} \operatorname{sgn}(\tau)\, \alpha(v_{\tau(1)}, \dots, v_{\tau(p)})\, \beta(v_{\tau(p+1)}, \dots, v_{\tau(p+q)}). \end{align*} This is the claimed identity. Since $v_1, \dots, v_{p+q} \in V$ were arbitrary, the shuffle formula holds as an identity of alternating multilinear forms. $\blacksquare$ [/step]