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eventually, is 0 even or it is neither even nor odd?
Zero is an even integer. (Zero is neither positive nor negative).
An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).
Check out factorial explanation for those who are confused. This is an explanation from someone else that I liked and makes sense.
You cannot reason that x^0 = 1 by thinking of the meaning of powers as "repeated multiplications" because you cannot multiply x zero times. Similarly, you cannot reason out 0! just in terms of the meaning of factorial because you cannot multiply all the numbers from zero down to 1 to get 1.
Mathematicians *define* x^0 = 1 in order to make the laws of exponents work even when the exponents can no longer be thought of as repeated multiplication. For example, (x^3)(x^5) = x^8 because you can add exponents. In the same way (x^0)(x^2) should be equal to x^2 by adding exponents. But that means that x^0 must be 1 because when you multiply x^2 by it, the result is still x^2. Only x^0 = 1 makes sense here.
In the same way, when thinking about combinations we can derive a formula for "the number of ways of choosing k things from a collection of n things." The formula to count out such problems is n!/k!(n-k)!. For example, the number of handshakes that occur when everybody in a group of 5 people shakes hands can be computed using n = 5 (five people) and k = 2 (2 people per handshake) in this formula. (So the answer is 5!/(2! 3!) = 10).
Now suppose that there are 2 people and "everybody shakes hands with everybody else." Obviously there is only one handshake. But what happens if we put n = 2 (2 people) and k = 2 (2 people per handshake) in the formula? We get 2! / (2! 0!). This is 2/(2 x), where x is the value of 0!. The fraction reduces to 1/x, which must equal 1 since there is only 1 handshake. The only value of 0! that makes sense here is 0! = 1.