Understanding the transfer function of an FIR filter I'm currently studying FIR filter and am having trouble understand how the following equation works, and it's implication.

$$ y[n] = h[n] * z^n = H(z) \cdot z^n $$

I don't really understand how this works given that

$$ y[n] = h[n] * x[n] $$
$$ Y(z) = H(z) \cdot X(z) $$

It seems to me that $ x[n] = X(z) = z^n $.  What's going on here?  What domain is $ z^n $ in?

I thought that convolution is done on two functions in the n-domain; however, in the equation $ y[n] = h[n] * z^n $, $ z^n $ does not appear to me to be in the n-domain.

If it is in the n-domain, then I don't understand how the other part works, $ y[n] =  H(z) \cdot z^n $, because I thought that then it should be

$$ y[n] = Z^{-1}(\,H(z) \cdot Z(z^n)\,) $$

Thanks.

Edit:
Here's the page from the book with the equation.  It does seem to me that $z^n$ is used in both the time and frequency domain.
Looks like a notation or transcription error. The "z" that's used as the transform variable to calculate the Z-Transform can not show up as an independent variable in the time domain as well. One of them needs a different name.