Why does $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$ have new divisors $59$ and $509$ all of a sudden? I am a noob when it comes to math so please bear with me. Why $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$ has $2$ new divisors $59$ and $509$. I mean, all of its divisors are prime factors and by no way $59$ nor $509$ can be reached when multiplying any of its divisors. 
How does the remainder $1$ affect the divisibility of a number?

You can understand this much better if you know the concept of 'co-primes'.
Two numbers are said to be co-primes, if they do not have common factor other than 1. 
Also note that for any positive integer $n$ where $(n\neq1)$, $n$ and $n+1$ are always co-primes as they will not have any common factor.Hence they are co-primes.
Let $a= 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$.
Let $b = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$
As $a$ and $b$ are co-primes, none of the factors which divide $a$ will divide $b$.
Also note that any number of the form $p_1 \cdot p_2 \cdot  \cdot  \cdot  \cdot p_n + 1$,where $p_1,p_2,\cdot\cdot\cdot p_n$ are all the primes till $p_n$, is either a prime or divisible by a prime not included in the list. 
