The Fourier transform of a product of $f$ with $t^n$ There is a proof in my book I don't quite follow.  We are supposed to prove that the Fourier transform of a product of $f$ with $t^n$ is given by
$$\mathcal{F}[t^n f(t)] (\lambda) = i^n \frac{d^n}{d \lambda^n} \{\mathcal{F} [f] (\lambda)\}$$
I will show the proof up until the point I don't get.  The proof is:
For the Fourier transform of a product of $f$ with $t^n$, we have
$$\mathcal{F}[t^n f(t)] (\lambda) = \frac{1}{2 \pi} \int_{- \infty}^{\infty} t^n f(t) e^{- i \lambda t} dt$$
Using
$$t^n f(t) e^{-i \lambda t} = (i)^n \frac{d^n}{d \lambda^n} \{f(t) e^{-i \lambda t} \}$$, 
we obtain. . .
Now the rest of the proof I understand.  However, I don't see where the book gets the expression
$$t^n f(t) e^{-i \lambda t} = (i)^n \frac{d^n}{d \lambda^n} \{f(t) e^{-i \lambda t} \}$$
from.  If anyone can explain this to me, I would be very grateful!

We have:
$$\frac{d}{d \lambda} \{f(t) e^{-i \lambda t} \} = (-it) f(t) e^{- i \lambda t}$$
Differentiating again gives:
$$\frac{d^2}{d \lambda ^2} \{ f(t) e^{- i \lambda t} \} = (-it)^2 f(t) e^{- i \lambda t}$$
So in general:
$$\frac{d^n}{d \lambda ^n} \{ f(t) e^{- i \lambda t} \} = (-it)^n f(t) e^{- i \lambda t}$$
It is now an easy task to check that the two sides of the equation stated in the problem are equal.
