Can the Product of n Consecutive Positive Integers Be Divisible by n!?

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Homework Statement


Using the minimal counterexample technique prove that the product of n consecutive positive integers is always divisible by n!

The Attempt at a Solution



Suppose that the statement is not true and the product of n consecutive positive integers is not always divisible by n!...

I'm stuck here. How can I continue this to get a contradiction and thus show that the product of is always divisible by n! ? Any guidance is greatly appreciated.
 
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Then there is a least counterexample k for which the proposition does not hold. Since this is the least counterexample, the proposition holds for k-1. Now you need to do some algebraic tricks to show that the proposition holding for k-1 implies that it holds for k. Your set up should look something like
c*(k-1)!= a * (a+1) *...*(n+a-1), for some a,c>0
 
JThompson said:
Then there is a least counterexample k for which the proposition does not hold. Since this is the least counterexample, the proposition holds for k-1. Now you need to do some algebraic tricks to show that the proposition holding for k-1 implies that it holds for k. Your set up should look something like
c*(k-1)!= a * (a+1) *...*(n+a-1), for some a,c>0

What does that "c" mean? Could you please explain a bit more how you got this expression?
 
Note: Sorry- the last term in the equation I posted had an error- the last term should have been a+k-2.

Definition of "divides": For integers x and y, where x does not equal 0, we say x divides y if there exists an integer c so that y=xc.

So in the equation c*(k-1)!= a * (a+1) *...*(a+k-2), the terms a, a+1, a+2 ... a+k-2 are the consecutive integers (note that there are a+k-2 - a +1 = k-1 of them), and c is the integer from the definition of integer divisibility above. This is the inductive assumption.

Edit: Forget the part about the inductive assumption- different type of proof. Forgot this one used least counterexample.