You're a little bit off with your definition of transcendental numbers. The "algebraic numbers" (i.e. the "non-transcendental" numbers) can be thought of as follows. We begin by considering the numbers that can be brought to zero under the following operations:
- addition by an integer
- multiplication by a (nonzero) integer
- (positive) integer powers
Any such number and any sum of finitely many of these numbers is algebraic. If a number isn't algebraic, it's transcendental.
$3$ is algebraic because $3-3=0$.
$\frac34$ is algebraic because $4\times \frac34 = 3$, and $3-3=0$.
$\sqrt3$ is algebraic because $(\sqrt3)^2=3$, and $3-3=0$.
$e$ is transcendental because no amount or order of these operations will bring $e$ to $0$. We can't simply divide $e$ by $e$ or do anything like that because $e$ is not a rational number, and we can't call that one step because $e$ certainly is not an integer.
The equivalent (and more concise) statement of this definition, by the way, is that an algebraic number is one that is the root of some polynomial
$a_0+a_1x+a_2x^2+\dots+a_nx^n$, where each coefficient $a_k$ is an integer. So for example, $\sqrt2-1$ is algebraic because
$$
(\sqrt2-1) + 1=\sqrt 2\\
(\sqrt 2)^2=2\\
2-2=0
$$
Equivalently, $\sqrt 2-1$ is a root of the equation
$$
(x+1)^2-2=0\implies\\
x^2+2x-1=0
$$
Note: You could (as some do) add to the list of allowable operations division by an integer, which would allow you to consolidate some steps by using rational numbers. However, this does not expand the set of numbers you can bring to zero because any root of a polynomial with rational coefficients is the root of some polynomial with integer coefficients.
Along the same lines, you could allow negative/rational powers with out changing the outcome.
Second note: if you're trying to understand all this, I highly recommend this video (if you aren't already a numberphile fan), which does a good job of explaining what I've said here.
You've successfully proved the the Lemma, and then the Theorem. And should feel proud that you succeeded.
I post this "community post" in part, so that an answer exists, and is posted objectively, (no one has anything to gain), hence removing one potential "unanswered" questions.
Best Answer
The book is wrong. If, for instance, you raise $1+\sqrt2$ to some power $n$; you never get a rational number. Every number of the form $\left(1+\sqrt2\right)^n$ can be written as $a+b\sqrt2$, with $a,b\in\Bbb N(=\{1,2,3,\ldots\})$, and it is therefore irrational.