youll need an outer loop which loops as high as the amount of factorials you want to compute, and an inner loop which will compute the factorial at the outer loops number.

pseudocode:

input n
set x = 1
loop n times
set num = 1
set y = 0
loop from x to 1
multiply num by y and re-store it into num
end loop
print num
end loop

that should do the job....unless i overlooked something

Well, first of all you need to know what a factorial is. You haven't demonstrated that you know anything on how to solve this problem yet. If you need a description on what a factorial is here's a link. http://mathworld.wolfram.com/Factorial.html
You would have to have a loop inside of a loop in order to make this work. Here's a little bit of code to get you started on your merry way:

int factorial = 4;
int result = 1;
for (int i = 0; i < factorial; i++)
result *= i+1;
System.Console.WriteLine("result:" + result);

the factorial is saved to the int result. If you want multiple answers just put the whole thing in a for loop incrementing factorial till you get to the desired value.
Hope this is helpful. I don't want to solve the whole thing for you though...

thanks a lot for the help both of ya guys..the way i did using the code above as following and worked perfectly fine for the results required...

static void Main(string[] args)
{
int num;
int j;
Console.WriteLine("Enter the number: ");
num = int.Parse(Console.ReadLine());
for (int i = 1; i <=num; i++)
{
Console.Write("1");
int total = 1;
for (j=1; j <= i; j++)
{
Console.Write("X" + j);
total = total*j;
}
Console.WriteLine(" = " + total);
}
}

That looks great. I'm glad I could help you. I just took a math assessment and I needed a little encouragment. lol. Don't hesitate to ask if you need help on anything else.

Thanks for that response. That's what I was going to say. This problem does require a little extra thinking for really large numbers. You need to have an understanding about basic programming.

public class Main {
public static void main(String[] args) {
long timeCheck;
long periodOfTime;
ScheduledExecutorService scheduler = Executors.newScheduledThreadPool(2);
long initialDelay = 0; // ...