Every binomial polynomial is irreducible but not prime in the ring of integer-valued polynomials over rational integers

Statement
In the ring of integer-valued polynomials over rational integers, every binomial polynomial, i.e., every polynomial of the form:

$$\binom{x}{r}$$

is an fact about::irreducible element but not a fact about::prime element.

The binomial polynomial is irreducible
For this, we need a lemma: the largest number that we can guarantee divides the value of any polynomial of the form $$\prod_{i=1}^r (x - a_i)$$ for all $$x \in \mathbb{Z}$$ is $$r!$$.

With this lemma, we observe that if $$\binom{x}{r}$$ is expressed as a product of polynomials of smaller degree, each of them is of the form $$(p_i/q_i)$$ times a product of linear polynomials. In each of the cases, we get that $$q_i$$ is bounded from above by the factorial of the number of linear polynomials in that factor. Thus, the product of all the $$q_i$$s is bounded by the product of the factorials, which is strictly less than $$r!$$, a contradiction.

The binomial polynomial is not prime
For $$r > 1$$, we have:

$$r!\binom{x}{r} = x(x-1)(x-2) \dots (x-r + 1)$$.

This yields:

$$\binom{x}{r} | x(x-1)(x-2) \dots (x-r+1)$$.

On the other hand, we have that since $$r > 1$$, $$\binom{x}{r}$$ does not divide any of the linear polynomials $$x - i$$.

For $$r = 1$$, use that:

$$x | x(x-1) = 2\binom{x}{2}$$

but $$x$$ does not divide $$\binom{x}{2}$$, since the ratio, $$(x-1)/2$$, is not an integer-valued polynomial.