Proof plan. The estimate follows directly from the explicit integral kernel of the Schrödinger propagator, which is a rescaled complex Gaussian. Taking the absolute value under the integral and using $|e^{i\alpha}| = 1$ for $\alpha \in \mathbb{R}$ gives the bound immediately.

Step 1: The explicit kernel. By completing the square in the Fourier inversion of $e^{-it|\xi|^2}\hat{u}_{\mathrm{in}}(\xi)$ (see e.g. the heat kernel calculation with $t$ replaced by $it$), the propagator has the integral representation
\begin{align*} (e^{it\Delta}u_{\mathrm{in}})(x) = \frac{1}{(4\pi it)^{d/2}} \int_{\mathbb{R}^d} e^{i|x-y|^2/(4t)}\, u_{\mathrm{in}}(y)\, d\mathcal{L}^d(y), \end{align*}
where $(4\pi it)^{d/2} = (4\pi|t|)^{d/2} e^{id\pi/4 \operatorname{sgn}(t)}$.

Step 2: The pointwise bound. For every $x \in \mathbb{R}^d$ and $t \ne 0$:
\begin{align*} |(e^{it\Delta}u_{\mathrm{in}})(x)| &= \frac{1}{(4\pi|t|)^{d/2}} \left|\int_{\mathbb{R}^d} e^{i|x-y|^2/(4t)}\, u_{\mathrm{in}}(y)\, d\mathcal{L}^d(y)\right| \\ &\le \frac{1}{(4\pi|t|)^{d/2}} \int_{\mathbb{R}^d} \underbrace{|e^{i|x-y|^2/(4t)}|}_{= 1}\, |u_{\mathrm{in}}(y)|\, d\mathcal{L}^d(y) \\ &= \frac{1}{(4\pi|t|)^{d/2}} \|u_{\mathrm{in}}\|_{L^1(\mathbb{R}^d)}. \end{align*}
Taking the supremum over $x$ gives the result with $C_d = (4\pi)^{-d/2}$.