Is $\frac{F[x_1, x_2, \ldots, x_n]}{<x_n>} \cong F[x_1, x_2, \ldots, x_{n-1}]$?

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I'm currently studying for my abstract algebra exam and came along this question in one of the reviews:

If F[$x_1, x_2, \ldots, x_n$] is PID, then F must be a field

I know that if F[x] is a PID, then F is a field. So if it's true that $\frac{F[x_1, x_2, \ldots, x_n]}{<x_n>} \cong F[x_1, x_2, \ldots, x_{n-1}]$, I could do an induction argument to do the proof. I just don't know if this statement is true (although it does seem to true). I've tried to come up with a ring homomorphism:

$\phi: {F[x_1, x_2, \ldots, x_n]} \rightarrow F[x_1, x_2, \ldots, x_{n-1}]$

where ker $\phi = <x_n>$ so I could apply the first isomorphism theorem, but haven't had any luck.

Thank you!