I want to show that $$\mathcal{F}[\psi^{(n)}(x)] = \left(i\frac{p}{\hbar}\right)^n\bar{\psi}(p)$$ where
$$\mathcal{F}[\psi(x)] = \bar{\psi}(p) = \frac{1}{\sqrt{2\pi\hbar}}\int^\infty_{-\infty} e^{i\frac{px}{\hbar}}\psi(x) dx$$
by taking:
$$\bar{\psi}(p) = \frac{1}{\sqrt{2\pi\hbar}}\int^\infty_{-\infty}e^{i\frac{px}{\hbar}}\psi(x) dx$$
And integrating by parts:
$$\frac{1}{\sqrt{2\pi\hbar}}\int^\infty_{-\infty}dxe^{i\frac{px}{\hbar}}\psi(x) = -\frac{i\hbar}{p}e^{i\frac{-px}{\hbar}}\psi(x)\bigg|^{\infty}_{-\infty} + \int^{\infty}_{-\infty}dx\frac{i\hbar}{p}e^{i\frac{-px}{\hbar}}\psi^{(1)}(x) = \\ = \sum^{1}_{j=0} - \frac{i\hbar}{p}e^{i\frac{-px}{\hbar}}\psi^{(j)}(x)\bigg|^{\infty}_{-\infty} + \int^{\infty}_{-\infty}dx\big(\frac{i\hbar}{p}\big)^2e^{i\frac{-px}{\hbar}}\psi^{(2)}(x)$$
If I do it $n$ times:
$$\bar{\psi}(p) = \sum^{n-1}_{j=0} - \frac{i\hbar}{p}e^{i\frac{-px}{\hbar}}\psi^{(j)}(x)\bigg|^{\infty}_{-\infty} + \int^{\infty}_{-\infty}dx\big(\frac{i\hbar}{p}\big)^ne^{i\frac{-px}{\hbar}}\psi^{(n)}(x),$$
If the sum goes to zero I think I can reach the relation, but I dont know why it should go to zero, but I still have a problem with the sing of i when the $\left(\frac{i\hbar}{p}\right)^n$ goes to the left-hand side.
So I want to ask is:
is possible to prove that relation in that way?
if it is, which arguments should I use to say that de summation goes to zero?
Thanks in advance !!