"Canceling Out The Zeroes" In A Mathematically Sane Way $\frac{0\times x}{0\times 1}$

Question

Introduction

Lets look at the product sequence: $$(n-1)(n-2)(n-3)...(n-k)$$

Where $n,k\in \mathbb N$ and $n\le k$ ; the expression will always have value $0$

But what if we remove the $n$th term in the sequence for choosen $n$?
How would you then calculate the value?

For example, we first pre-select $k$, lets say 7, then we have:

$(n-1)(n-2)...(n-7)$

Now choose $n$ lets say $4$ and remove the $4$th term;

$(4-1)(4-2)(4-3)(\text{"removed"})(4-5)(4-6)(4-7)=$
$3\times 2 \times 1 \times ("0") \times -1 \times -2 \times -3=$
$3!\times -(3!)=$
$-36$

I have found a formula for $n\lt k $ like: $$ (k-n)!\times (-1)^{k-1} \times(n-1)!$$

For $n=k$ it is: $$(n-1)!$$

And for $n\gt k$ it would be: $$\frac{(n-1)!}{(n-k-1)!}$$

---

Problem

How would you remove the $n$th term mathematically?

One way of doing so is using the sets and $\prod$ as described below, but could it be done pure calculus/algebra as a single expression uniting all 3 conditions for $k$ ?

---

My instant first thought was writing the whole thing like: $$\frac{(n-1)...(n-a)...(n-k)}{(n-a)}$$

But in this form, we are always choosing $n=a$ and we can't cancel out the two zeroes since that is not how mathematical operations work and dividing by zero isn't defined.

I was thinking then using the sets, like defining the set $\mathbb M$ with items: $m_1, m_2, m_3... = (n-1), (n-2), (n-3)...$

Then doing $\mathbb M \setminus (n-a)$ where we choose legal $a$.

And finally using: $$\prod_{m\in M}m$$

Which notates multiplication of all items that are left in a set, for $n=a$ (Choosing the same $n$ as our previously chosen $a$)

---

In other words

We write the expression as: $$\frac{(n-1)...(n-a)...(n-k)}{(n-b)}$$

Then we instruct to choose variables as it follows:

1) Choose $k$ to define the finite length of the sequence.

2) Choose a number $x \in \mathbb N$ for $x=a=b$ and cancel out the two.

3) Finally choose $n=x$ and compute as already stated in introduction...

Again, how would one write this in a more mathematical way? (As a single expression possibly for all 3 conditions given in introduction, without needing the step by step explanation)

Following the steps, We are ignoring the condition needed in step 2: $(n-x)\ne 0$

---

In Addition

Reading the comments makes it more clear, sorry for any inconveniences.

This thought came to me while I was experimenting with this: "Binary-Like" Function?; In Consecutive Products as Multi-Factorials....

Answer 1

I should start out by saying that the answer to your question— as asked— is yes: you have already done it. By "done it" I mean "removed the $n^\text{th}$ term; your calculation in the introduction is precisely how you remove the $n^\text{th}$ term.

The question I suspect you are trying to ask, does not have an answer, because the question is not well-formulated; something like "Is there some one-line expression I can write down to give me these numbers?".

A stupid answer is "yes, trivially", since you can just define $F(n,k)$ to be the function that outputs the numbers you need it to be, and then of course you can just write down "$F(n,k)$" on the paper and this is the answer to your question.

The moral lesson we learn from the stupid answer is that:

This issue critically depends on what expressions you are letting yourself use.

And you have to be extremely nitpicky about this. For instance, if you let yourself use the $\prod$-schema in general, you already have your answer in the question. If you don't allow generic sets in your limits, you could still write

$$\prod_{\substack{1\leq i\leq k \\ i\neq n}} (n-i)$$

Perhaps you are only allowed to use products over intervals? Then you could write

$$\prod_{i=1}^{n-1} (n-i) \cdot \prod_{i=n+1}^k (n-i)$$

But this is maybe not satisfying because in order to make sense of it for $k\leq n$, you have to be willing to work with products over empty intervals.

(By the way, once you specify what expressions you are letting yourself use, you have now made the problem well-defined. But you're not out of the woods yet: showing the answer is 'no' is likely to be very difficult. And showing the answer is 'yes' is theoretically just as easy as finding the expression... but in practice this can be virtually impossible.)

Answer 2

What you want is a product of falling factorials. With the factor $(n-j)$ removed,

$$P(n,k,j)=(n-1)_{j-1}(n-j-1)_{k-j}.$$

---

When you specialize the formula with $j=n$, you indeed get

$$P(n,k,n)=(n-1)_{n-1}(-1)_{k-n}=(n-1)!(-1)^{k-n}(k-n)!$$

When $n>k$, this is undefined, just like with the initial definition. (Notice that by analytical continuation, the factorial of a negative is $\pm\infty$.)