How can one rewrite $x(x-1)\cdots(x-k+1)$ for $k = 0$? While trying to prove one expression via mathematical induction, I've came to the dead end – I can't deduce, which form will sequences $x(x-1)\cdots(x-k+1)$ and $(x-1)(x-2)\cdots(x-k)$ take for k = 0. How can I find it?
 A: The notation $x(x-1) \cdots (x-n+1)$ is ambiguous when $n=0$.  A text defining this product should give a separate definition for this case.  For example, Wikipedia writes 

The value of each is taken to be 1 (an empty product) when n=0.

Since the formula is ambiguous, in some sense it's a matter of choice how to define it when $n=0$.  But there are a number of good reasons why it's more convenient to set it to $1$.  This is similar to the situation with something like $x^n$.  If you write your own textbook, I suppose you could define $x^n = 7$ when $n = 0$, but your formulas would quickly become very ugly with a lot of special cases when $n=0$.
Here are some good reasons why $x(x-1) \cdots (x-n+1)$ should be $1$ when $n =0$:


*

*It's an example of an empty product which is generally taken to be $1$ for many reasons (see the link.)

*We would like to write $x(x-1) \cdots (x-n+1) = n!{ x \choose n}$, and we have $n! = {x \choose n} = 1$ when $n=0$, again because they are empty products.

*There's a combinatorial interpretation for $x(x-1) \cdots (x-n+1)$: it's the number of ways to choose $n$ labeled balls from a bucket containing $x$ labeled balls, without replacement, in some order.  If $n=0$ there's exactly one way to choose them: don't choose any. 

A: If the goal is to prove something by mathematical induction, an alternative is to let $k=1$ be the base case rather than $k=0$.  Then (assuming you can prove the induction step) you can conclude that the claim is true for all $k \ge 1$.
