[proofplan] We first compute the effect of one coordinate derivative on the exponential phase. Multiplication by $-i$ converts the derivative factor $i(y_j-x_j)$ into the coordinate factor $y_j-x_j$. Iterating this coordinate identity according to the multi-index $\alpha$ and using that the coordinate factors are independent of $\eta$ gives the desired product formula. [/proofplan]

[step:Compute the one-coordinate differentiated exponential]Define the smooth function \begin{align*} \Phi: \mathbb{R}^n \to \mathbb{C}, \qquad \eta \mapsto e^{i(y-x)\cdot \eta}. \end{align*} For each $j \in \{1,\dots,n\}$, the chain rule applied to the real-linear phase map $\eta \mapsto (y-x)\cdot \eta$ gives \begin{align*} \partial_{\eta_j}\Phi(\eta) = i(y_j-x_j)e^{i(y-x)\cdot \eta}. \end{align*} Multiplying by $-i$ yields \begin{align*} (-i)\partial_{\eta_j}\Phi(\eta) = (y_j-x_j)e^{i(y-x)\cdot \eta}. \end{align*}[/step]

[guided]Define the function being differentiated by \begin{align*} \Phi: \mathbb{R}^n \to \mathbb{C}, \qquad \eta \mapsto e^{i(y-x)\cdot \eta}. \end{align*} The variables $x$ and $y$ are fixed parameters in this differentiation, while $\eta$ is the variable. For a fixed coordinate $j \in \{1,\dots,n\}$, the phase function is \begin{align*} \eta \mapsto (y-x)\cdot \eta = \sum_{k=1}^n (y_k-x_k)\eta_k. \end{align*} Its $\eta_j$-derivative is the constant coefficient $y_j-x_j$. Applying the chain rule to the exponential map gives \begin{align*} \partial_{\eta_j}\Phi(\eta) = i(y_j-x_j)e^{i(y-x)\cdot \eta}. \end{align*} The convention for the Fourier differential operator includes the factor $-i$ precisely to cancel the $i$ produced by differentiating the oscillatory exponential. Thus \begin{align*} (-i)\partial_{\eta_j}\Phi(\eta) = (-i)i(y_j-x_j)e^{i(y-x)\cdot \eta} = (y_j-x_j)e^{i(y-x)\cdot \eta}. \end{align*}[/guided]

[step:Iterate the coordinate identity according to the multi-index] For each $j \in \{1,\dots,n\}$, the factor $y_j-x_j$ is independent of $\eta$. Repeated application of the coordinate identity therefore gives, for every $m \in \mathbb{N}_0$, \begin{align*} (-i)^m \partial_{\eta_j}^m \Phi(\eta) = (y_j-x_j)^m \Phi(\eta). \end{align*} Applying this with $m=\alpha_j$ in each coordinate and using the commutativity of mixed partial derivatives for the smooth function $\Phi$ gives \begin{align*} (-i)^{|\alpha|}\partial_\eta^\alpha \Phi(\eta) = \prod_{j=1}^n (y_j-x_j)^{\alpha_j}\Phi(\eta). \end{align*} By the definitions of $D_\eta^\alpha$ and $(y-x)^\alpha$, this is exactly \begin{align*} D_\eta^\alpha e^{i(y-x)\cdot \eta} = (y-x)^\alpha e^{i(y-x)\cdot \eta}. \end{align*} [/step]

[step:Check the zero multi-index convention] If $\alpha=(0,\dots,0)$, then $|\alpha|=0$, $\partial_\eta^\alpha$ is the identity operator, and $(y-x)^\alpha=1$. Hence the asserted identity becomes \begin{align*} e^{i(y-x)\cdot \eta}=e^{i(y-x)\cdot \eta}. \end{align*} This agrees with the formula proved above and completes the proof. [/step]