Denote by $S^m$ the set of functions $p: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{C}$ such that $p \in C^{\infty}(\mathbb{R}^n \times \mathbb{R}^n)$ and $$\left| \frac{\partial^{\alpha+\beta}}{\partial \xi^{\alpha} \partial x^{\beta}} p(x,\xi) \right| \leq c_{\alpha,\beta} \cdot (1+|\xi|)^{m-|\alpha|} \tag{1}$$ for $\alpha,\beta \in \mathbb{N}_0^n$. $p$ is called the symbol of the pseudo-differential operator $$p(x,D)u(x) := (2\pi)^{-n} \cdot \int e^{\imath \, x \cdot \xi} \cdot p(x,\xi) \cdot \hat{u}(\xi) \, d\xi$$ where $u \in \mathcal{S}$ is a Schwartz function, $\hat{u}$ denotes the Fourier-transform of $u$.
Theorem For $p \in S^{m_1}, q \in S^{m_2}$ we have that the composition $(p(\cdot,D) \circ q(\cdot,D))$ is again a pseudo-differential operator, with symbol in $S^{m_1+m_2}$.
Using the definition of the composition of operators, one can show that $$\begin{align} t(x,\xi) &:=e^{-\imath \, x \cdot \xi} \cdot p(x,D)[q(x,\xi) \cdot e^{\imath \, x \cdot \xi}] \\ &=\int k(x,x-y) \cdot q(y,\xi) \cdot e^{\imath \, (y-x) \cdot \xi} \, dy \end{align}$$ is the symbol of the operator $(p(\cdot,D) \circ q(\cdot,D))$ where $k(x,\cdot)$ denotes the inverse Fourier-transform (in the sense of distributions) of the mapping $\xi \mapsto p(x,\xi)$. So, I have to show that $t$ satisfies $(1)$ for $m:=m_1+m_2$. But I don't see how to proceed...
Any hints would be appreciated.