Meaning of "for sufficiently large" in infinite products The book I'm reading states that any positive integer $a$ greater than 1 can be expressed as a product of primes,
$$a=\prod_p{p^{\alpha{(p)}}}$$ where $\alpha{(p)}$ is a non-negative integer.
And that it is understood for sufficiently large primes $p$, $\alpha{(p)}=0$.
My question is: what is considered to be a large prime? And how can the statement $\alpha{(p)}=0$ for large primes $p$ be true? Does that mean large primes can never be factors of any integers? If so I find this very unintuitive.
 A: What that means is that, for each $n>1$, you can write $n$ as$$2^{\alpha(2)}\times3^{\alpha(3)}\times5^{\alpha(5)}\times7^{\alpha(7)}\times\cdots,$$where $\alpha(p)=0$ if $p$ is sufficiently large; in other words, you only have $\alpha(p)\ne0$ for finitely many primes.
Note that here “sufficiently large” depends on $n$. For $n=10$, “sufficiently large” means $p>5$, whereas for $n=74$, it means $p>37$.
A: 
Does that mean large primes can never be factors of any integers?

Of course not: each prime is a factor of itself.
What's sufficiently large depends on $a$. it means for each $a$, there is some $N$ for which any $p>N$ is sufficiently large. We can take $N$ to be $a$'s greatest prime factor.
A: This is just another way of saying that for all but finitely many primes $p$, we have $\alpha(p) = 0$. In particular, the "largeness of the prime" depends upon the $a$ you are given to start with.
Specifically how this relates to your case. You are given $a \geq 1$, then, for the most unrefined bound, $p > a$ can never be factors of $a$.
A: Here we formally tweak the books's statement of the theorem to add some insight.
Let $P$ denote the set of all prime numbers. Clearly $P$ is a well-ordered set.
If $a$ is an integer and $a \ge 2$ then any prime factor (and at least one exists) of $a$ must be less than or equal to $a$. Said another way, we can associate to $a$ a prime factor ${p^{max}_a}$ satisfying
$\tag 1 [\; {p^{max}_a}  \mid a \;] \land [(p \in P) \land (p \mid a) \implies p \le {p^{max}_a}]$
You can easily prove $\text{(1)}$ without the FTA.
Here is the fundamental theorem of arithmetic:
For every integer $a \ge 2$ there exist there exist one and only function
$\quad \alpha: P \setminus \{p \in P \mid p \gt {p^{max}_a} \}  \to \Bbb N$
satisfying the following property
$\tag 2 \displaystyle a=\prod_{p \in \text{domain}(\alpha)} {p^{\alpha{(p)}}}$
A: This is equivalent to say that, for any fixed $a$, the product is for a finite number of terms
$$a=\prod_{p=p_1}^{p_N}{p^{\alpha{(p)}}}$$
with $p_N \le a$.
