Question

please prove that 0!=1

Answer 1

Its not provable. When you define the factorial operation you have:
\[n!=n(n-1)(n-2)...(2)(1); 0!=1\]
Its "un-provable" basically

Answer 2

was only wondering how it was concluded...

Answer 3

thnks for the effort though

Answer 4

Its not "concluded". Its just defined that way. Just as we define things like prime numbers. The only reason 0!=1 is because its convenient.For example consider the series:
\[\sum_{n=0}^{\infty}\frac{x^n}{n!}=\frac{x^0}{0!}+\frac{x}{1!}+...\frac{x^n}{n!}+...=e^x\]
Notice in the first term of the series you have 0! in the denominator. So defining 0!=0 would make the series that converges to e^x to fall apart.
Convenience.

Answer 5

That isn't the only series expansion that hinges on that fact. There are several more.

Answer 6

..that clears that up...