Theorem to Prove $e$ is Irrational

Question

There is a theorem in this textbook I am using and the theorem is the following:

If $e^{x}=1+x+\frac{x^{2}}{2!}+\cdots +\frac{x^{n}}{n!}+\cdots $, then the number $e$ is irrational.

Then the proof of this theorem goes on about proving that $e^{-1}$ is irrational.

I don't understand how the proof relates to the theorem. Is the theorem missing some information that lets me know that showing $e^{-1}$ is irrational is equivalent to showing $e$ is irrational?

Edit: Sorry about that, what I meant was I don't understand what the theorem is saying. I already understand the proof. :)

Answer 1

If a number is irrational then its reciprocal is irrational. For if its reciprocal could be written $a/b$ then the number itself could be written $b/a.$

Answer 2

The reciprocal of a non-zero rational number is also rational.

They proved $1/e$ is irrational then the conclusion is that $e$ is also irrational because otherwise its reciprocal $1/e$ would have been rational as well.