I don't quite understand "by the properties of the Fourier transform, $it^r\psi(t)$ is the Fourier transform of $(-1)^rD^r\Psi(x)$" from Wikipedia about Gram–Charlier A series and Edgeworth expansion:
We examine a continuous random variable. Let \mathit{f} be the characteristic function of its distribution whose density function is $\mathit{F}$, and $\kappa_r$ its cumulants. We expand in terms of a known distribution with probability density function $\Psi$, characteristic function $\psi$, and cumulants $\gamma_r$. The density $\Psi$ is generally chosen to be that of the normal distribution, but other choices are possible as well. By the definition of the cumulants, we have the following formal identity: $$ f(t)=\exp\left[\sum_{r=1}^\infty(\kappa_r-\gamma_r)\frac{(it)^r}{r!}\right]\psi(t)\,.$$ By the properties of the Fourier transform, $(it)^r\psi(t)$ is the Fourier transform of $(-1)^rD^r\Psi(x)$, where $D$ is the differential operator with respect to $x$. Thus, we find for F the formal expansion $$ F(x) = \exp\left[\sum_{r=1}^\infty(\kappa_r - \gamma_r)\frac{(-D)^r}{r!}\right]\Psi(x)\,. $$
In the equation for $f(t)$, $(it)^r$ is inside the exponential and $\psi(t)$ multiplies with the exponential, so how can $(it)^r$ and $\psi(t)$ multiply together to get $(it)^r\psi(t)$?
