Is j^{-p}=e^{-j\frac{p\pi}{2}} a valid exponential function?

[yungman]

Homework Statement

I want to verify
j^{-p}=e^{-j\frac{p\pi}{2}}

Homework Equations

e^{j\frac{\pi}{2}}=\cos(\frac{\pi}{2})+j\sin(\frac{\pi}{2})=j

The Attempt at a Solution

j^{-p}=(e^{j\frac{\pi}{2}})^{-p}=e^{-j\frac{p\pi}{2}}

Am I correct?
Thanks

 

[dirk_mec1]

Yes.