prime numbers and expressing non-prime numbers

What is the sum of no numbers at all? Zero, of course, since it is the additive identity: $x + 0 = 0$, where $x \neq 0$, or even if it is.

Now, what is the product of no numbers at all? It can't be zero, since, maintaining the stipulation that $x \neq 0$, we have $x \times 0 = 0$, and we said $x \neq 0$. The multiplicative identity is $1$, since $x \times 1 = 1$.

Hence, the product of no primes at all is $1$. The fundamental theorem of arithmetic is a subtlety that's unnecessary for answering your question.

This is mainly just an extended comment on Peter Foreman's answer. The (relatively difficult) uniqueness aspect of the Fundamental Theorem of Arithmetic is not needed for the OP's question, just the (easier) existence aspect.

What's missing from the OP's textbook is the qualifier in the correct assertion that every non-prime number greater than $1$ can be expressed as a product of primes. This is the existence aspect of FTA, and it can be proved by strong induction: If $n\gt1$ is not a prime, then $n=ab$ for some pair of integers with $1\lt a,b$. Both $a$ and $b$ must be less than $n$ (otherwise their product would be more than $n$), so we can assume, by strong induction, that each of them can be written as a product of primes, hence so can their product, which is $n$.

Remark: "Strong" induction means that you don't just assume an assertion is true for $n-1$ and then prove it for $n$, you assume it's true for all positive integers $k\lt n$. In this case the assertion is "if $k\gt1$ and $k$ is non-prime, then $k$ can be written as a product of primes." Note that the base case, $k=1$, is vacuously true, because $1$ is not greater than $1$.