Why can't a polynomial of degree $n$ have less than $n$ (complex) roots, counted with multiplicity? According to the fundamental theorem of algebra, any polynomial of degree $n$ has exactly $n$ (complex) roots counted with multiplicity)
My text book explains that an $n$ degree polynomial can have, at most, $n$ roots as follows:

If a $n$ degree polynomial had roots [ $α_1,α_2,...α_{n+1}$], then, by the factor theorem, each of $(x-α_1),(x-α_2)...(x-α_{n+1})$ must be a factor of P(x), and so must be their product $$\Rightarrow P(x)=(x-\alpha_1)(x-\alpha_2)...(x-\alpha_{n+1})\ \cdot\ Q(x)\quad \text{where Q(x) is another polynomial}$$
But that would mean that the degree of the right is at least $n+1$, so P(x) cannot be an $n$ degree polynomial.

To me, this only proves that if a polynomial has $n$ roots, it must be at least $n$ degree. What I don't understand is how this implies that there can't be an $n$ degree polynomial with less than $n$ roots (even when counted with multiplicity).
Following the same logic used in the text book, if there is a polynomial of degree $n$ with $m<n$ known roots, P(x) can be expressed as $$P(x)=(x-\alpha_1)(x-\alpha_2)...(x-\alpha_{m})\ \cdot\ Q(x)\quad \text{where Q(x) is a polynomial of degree}\ (n-m)$$
How do you we know that Q(x) must also have $(n-m)$ roots ? I feel like this is a very stupid and obvious question, but I can't really explain why for some reason?
 A: We would know by the fundamental theorem of algebra, applied to $Q$.
The form I know the FTA only says "at least one root", and then "exactly $n$ roots" is a corollary. Regardless, once you know any non-constant complex polynomial has at least one root, the proof that the number of roots is exactly the degree follows, for instance, by induction on the degree.
For a first-degree polynomial, it is obvious they have exactly a single root. (If you're not being too picky, non-zero constant polynomials could also serve as a base case here, but then the induction step for degree $1$ and for degree greater than $1$ would have to be different anyways. So degree one is the base case I choose.)
Fix some $k>1$, and assume that any polynomial of degree less than $k$ has exactly as many roots as its degree. For any polynomial $Q_k$ of degree $k$, the FTA tells us that it has at least one root $a$, so we get $Q_k(x) = (x-a)Q_{k-1}(x)$ for some polynomial $Q_{k-1}$ of degree $k-1$. But that means by the induction hypothesis that $Q_{k-1}$ has exactly $k-1$ roots, and multiplying by $(x-a)$ adds exactly one root. So $Q_k$ has exactly $k$ roots.
A: 
To me, this only proves that if a polynomial has n roots, it must be at least n degree.

Correct.

What I don't understand is how this implies that there can't be an n degree polynomial with less than n roots (even when counted with multiplicity).

It doesn't.

As other people have stated, the hard part is knowing that every nonconstant polynomial has at least one root, because then we can just keep factoring out roots until we get down to a constant.
However, showing that a non-constant polynomial has at least one root is not an elementary result, and cannot be proven without using at least a little calculus.  Moreover, the proofs that just use a little bit of calculus are very unenlightening; they're basically the sort of thing you can only come up with when you already know the answer.  To really understand the fundamental theorem of algebra you need to know some complex analysis, which is a course you'd typically take after completing an entire college calculus sequence and maybe a few more courses on top of that.
Since it sounds like your book is not at this level, it sounds like they're telling you that the fundamental theorem of algebra is true to allow you to see what it's useful for, but they're definitely not proving it.
(To give you a preview, the cleanest proof works by taking the complex plane, wrapping it up into a sphere and gluing it together by sticking on a point at infinity, and then noticing that your polynomial has a "pole of degree $n$" at this point and cannot be infinite anywhere else.  It follows that it has to have $n$ zeroes, counted with multiplicity, in order to "balance out" how quickly it goes to infinity.)
