In General: 
It is obvious that the term Prime are strictly related to numbers that can not have any other factors except for the number one and itself.
For Instance: 
  1. The number \(14\) is not Prime since it has more than two factors we discussed, i.e. the numbers \((1,2,7,14)\) are divisors for the number \(14\).
  2. On the other hand, the Prime number \(11\) has just two divisors which are \((1,14)\) only.   
But, according to this definition, the number \(1\) are an eligible candidate for this definition. So, what other definition mathematicians have related to ?

The general definition above is not sufficient enough for determining whether the number \(1\) is prime or not! 

However, the prime-factorization theorem have the answer for this ambiguity. It is also called: "The Fundamental Theorem of Arithmetic", and that theorem states:  
\(\forall M \in \mathbb{N} \) \(M\) can be written as a unique product of primes      

Discussion:     
  1.  The Number \(24\) has the unique product of primes representation which is: \(24 = 2\times 2\times 2\times 3 \). 
  2. The Prime Number \(17\) has the unique prime product representation which is the number \(17\) itself.
So, the last theorem will be violated if the number \(1\) considered to be prime, and here is a counter example:

Let \(C \in \mathbb{N}\) be a composite number which has the following unique n-primes product \( p_{1}\times  p_{2}\times p_{3}\times ... \times p_{n} \), i.e. \(C= p_{1}\times p_{2}\times p_{3}\times ... \times p_{n} \). Now, Assume that \(1\) is prime number, then the composite number \(C\) has infinite number of primes product representations as follow: 
\(C= p_{1}\times p_{2}\times p_{3}\times ... \times p_{n} \)
\(C= 1\times p_{1}\times p_{2}\times p_{3}\times ... \times p_{n} \)
\(C= 1\times 1\times p_{1}\times p_{2}\times p_{3}\times ... \times p_{n} \)
\(C= 1\times 1\times 1\times p_{1}\times p_{2}\times p_{3}\times ... \times p_{n} \) 
and so on ...
Then, No unique prime product factorization will be existed and the upper theorem will be violated.
Hence, \(1\) is not prime number.

Numerically consider the number \(30\) which has the following unique prime product: \( 2\times 3\times 5\). But, if the number \(1\) is prime then there will be no such uniqueness in prime product representation for each of the positive whole numbers.
And here is how it looks like if \(1\) considered to be prime:
\(30=  2\times 3\times 5 \)
\(30= 1\times 2\times 3\times 5\)
\(30= 1\times 1\times 2\times 3\times 5\)
\(30= 1\times 1\times 1\times 2\times 3\times 5\) and so on...
Thus, \(1\) is not prime number.