Zeros At The End Of Factorial N

/*
 * * Write a program that calculates for given N how many trailing zeros present at the end of the number N!. Examples:
    N = 10  N! = 3628800  2
    N = 20  N! = 2432902008176640000  4
    Does your program work for N = 50 000?
    Hint: The trailing zeros in N! are equal to the number of its prime divisors of value 5. Think why!
*/

using System;
using System.Numerics;

class ZerosAtTheEndOfFactorialN
{
    static void Main()
    {
        BigInteger numberN;
        BigInteger trailingZeros = 0;
        BigInteger divider = 5;
        string invalidInput = "Please enter a valid number greater than 0!" + Environment.NewLine;
        Console.WriteLine("Enter a value for N: ");
        while (!(BigInteger.TryParse(Console.ReadLine(), out numberN) && numberN > 0))
        {
            Console.WriteLine(invalidInput);
            Console.WriteLine("Enter a value for N: ");
        }

        for (int i = 5; i <= numberN*2; i = i * 5)
        {
            trailingZeros = trailingZeros +  (numberN / i);
        }
        Console.WriteLine("There are {0} trailing zeros in factorial of number {1}!" + Environment.NewLine, trailingZeros, numberN);
        Main();
    }
}