recurrence equation

Formally, you should use induction.
Want to show f(n) = n! for n>0:

f(1)=1=1! , f(2)=2=2! by definition.

Assume there is a k>1 such that f(k)=k!.
Then f(k+1) = f(k) x (k+1) = k! x (k+1) = (k+1)!.

Therefore, by induction, f(n) = n! for n>0.

Finished~~~

Note:
In the induction assumption we need to assume k>1 so that k+1>2
and hence the recurrence relation can be used.