Partons de la formule

$$\mathop{\mathrm{d}^{\prime}\mathrm{d}^{\prime\prime}}h=\sum_{i,j=1}^{n}\frac{\partial^{2}}{\partial x_{i}\partial x_{j}}\mathop{\mathrm{d^{\prime}}}x_{i}\mathop{\mathrm{d}^{\prime\prime}}x_{j}.$$

Alors, dans l’expression

$$(\mathop{\mathrm{d}^{\prime}\mathrm{d}^{\prime\prime}}h)^{n}=\sum_{\begin{subarray}{c}i_{1},\dots,i_{n}\\ j_{1},\dots,j_{n}\end{subarray}}\prod_{k=1}^{n}\frac{\partial^{2}}{\partial x_{i_{k}}\partial x_{j_{k}}}\mathop{\mathrm{d^{\prime}}}x_{i_{k}}\mathop{\mathrm{d}^{\prime\prime}}x_{j_{k}},$$

tous les termes pour lesquels $\{i_{1},\dots,i_{n}\}\neq\{1,\dots,n\}$ ou $\{j_{1},\dots,j_{n}\}\neq\{1,\dots,n\}$ sont nuls. Ainsi,

$$(\mathop{\mathrm{d}^{\prime}\mathrm{d}^{\prime\prime}}h)^{n}=\sum_{\sigma,\tau\in\mathfrak{S}_{n}}\prod_{k=1}^{n}\frac{\partial^{2}}{\partial x_{\sigma(k)}\partial x_{\tau(k)}}\,\mathop{\mathrm{d^{\prime}}}x_{\sigma(1)}\mathop{\mathrm{d}^{\prime\prime}}x_{\tau(1)}\dots\mathop{\mathrm{d^{\prime}}}x_{\sigma(n)}\mathop{\mathrm{d}^{\prime\prime}}x_{\tau(n)}.$$

Remarquons aussi que

$$\mathop{\mathrm{d^{\prime}}}x_{\sigma(1)}\mathop{\mathrm{d}^{\prime\prime}}x_{\tau(1)}\dots\mathop{\mathrm{d^{\prime}}}x_{\sigma(n)}\mathop{\mathrm{d}^{\prime\prime}}x_{\tau(n)}=\varepsilon(\sigma)\varepsilon(\tau)\mathop{\mathrm{d^{\prime}}}x_{1}\mathop{\mathrm{d}^{\prime\prime}}x_{1}\dots\mathop{\mathrm{d^{\prime}}}x_{n}\mathop{\mathrm{d}^{\prime\prime}}x_{n},$$

où $\varepsilon(\cdot)$ désigne la signature d’une permutation. En outre, si $\psi=\tau\circ\sigma^{-1}$, on a

$$\prod_{k=1}^{n}\frac{\partial^{2}}{\partial x_{\sigma(k)}\partial x_{\tau(k)}}=\prod_{k=1}^{n}\frac{\partial^{2}}{\partial x_{k}\partial x_{\psi(k)}}$$

et

$$\varepsilon(\sigma)\varepsilon(\tau)=\varepsilon(\psi).$$

Par suite,

$$(\mathop{\mathrm{d}^{\prime}\mathrm{d}^{\prime\prime}}h)^{n}=n!\sum_{\psi\in\mathfrak{S}_{n}}\prod_{k=1}^{n}\frac{\partial^{2}}{\partial x_{k}\partial x_{\psi(k)}}\varepsilon(\psi)\,\mathop{\mathrm{d^{\prime}}}x_{1}\mathop{\mathrm{d}^{\prime\prime}}x_{1}\dots\mathop{\mathrm{d^{\prime}}}x_{n}\mathop{\mathrm{d}^{\prime\prime}}x_{n},$$

ainsi qu’il fallait démontrer. ∎