Prove that every number between two numbersis composite

Question

I have this problem, and I have no idea how to go about it:

Let $p_1,p_2, \ldots, p_{n+1}$ be the first $n+1$ primes, in order. Prove that every number in between $p_1 \cdot p_2 \cdot p_3 \ldots p_n+1$ and $p_1 \cdot p_2 \cdot p_3 \ldots p_n+p_{n+1}-1$ (inclusive) is composite. How does this show that there are gaps of arbitrary length in the sequence of primes?

I've already viewed the other two questions with similar names, but neither have answers that make sense to me. I was hoping someone could show me step by step how to do this. I don't understand how it could be inclusive and still be true.

Answer 1

Something is wrong, at the very least you can't mean "inclusive," i.e. including the endpoints of the range of numbers, because if you put $n=1$ then the range of numbers you get is $3$ to $5$ (inclusive) and $3$ and $5$ are prime, although $4$ is composite so it might work if you remove the "inclusive" part.

Answer 2

I am assuming you only mean the "inclusive" to apply to the upper bound.

Every number $k$ from $2$ to $p_{n+1}-1$ has a prime factor $p$ (possibly $p$ is $k$ itself; if there is more than one possible $p$, just choose one). Moreover, since $k<p_{n+1}$, this factor $p$ cannot be $p_{n+1}$ or larger, but must be $p_1$ or $p_2$ or... or $p_n$. Therefore $$p\mid p_1p_2\cdots p_n$$ and so $$p\mid p_1p_2\cdots p_n+k\ .$$ Thus we have $p_{n+1}-2$ consecutive composite numbers (at least), and since there are infinitely many primes, this can be made arbitrarily large.