Need help on a proof involving the size of prime factors

Question

Prove that the prime factors of $510510^{510510} + 1$ are greater than or equal to 19.

Here is my (incomplete) proof that I need help with:
1. The prime factors of 510510 are 2, 3, 5, 7, 11, 13 and 17.
2. Since $510510^{510510}$ is a multiple of 510510, the prime factors of $510510^{510510}$ are also 2, 3, 5, 7, 11, 13 and 17.
3. ...

How can I show that none of these prime factors are factors of $510510^{510510} + 1$? In other words, how can I show that the smallest possible prime factor of $510510^{510510} + 1$ must be greater than the largest prime factor of $510510^{510510}$, which is 17?

Answer 1

Recall that consecutive integers are coprime. That is their gcd is 1. Thus no prime dividing the first can also divide the second.

More elementarily, you could argue: suppose $p$ divides both, then $p$ divides the difference, then $p$ divides $1$, a contradicition.