How to get $(n+1)^{\underline k} - n^{\underline k} = n^{\underline{k-1}}$?

Question

How I've understood it, a falling factorial, $n^{\underline k} = \frac{n!}{k!} = n(n-1)(n-2)...(n-k+1) = \displaystyle\prod_{i=k+1}^n i$

So, $$\Delta_n (n^{\underline k}) = (n+1)^{\underline k} - n^{\underline k} = \frac{(n+1)!}{k!} - \frac{n!}{k!} = \frac{(n+1)!-n!}{k!}$$

Let's ignore the denominator for now, for visual clarity's sake, and look at this difference.

$$(n+1)(n)(n-1)...(n-k+2)$$

minus

$$n(n-1)...(n-k+2)(n-k+1)$$

The top product has a factor that the bottom one doesn't have, and the bottom product has a factor that the top one doesn't have. This means this is a situation of $ab -bc$, where $a=(n+1)$, $b = n!$ and $c= (n-k+1)$. Now, $ab-bc = (a-c)b$, which means $(n+1)! -n! = (n+1-(n-k+1))n! = kn!$

Bringing back the denominator we put aside a moment earlier, this becomes $\displaystyle\frac{kn!}{k!} = kn^{\underline k} = n^{\underline{k-1}}$

This article however, puts the value at $kn^{\underline{k-1}}$, so where did I go wrong?

Answer 1

$n^{\underline k}=\frac{n!}{(n-k)!}$, not $\frac{n!}{k!}$, is your first mistake.

Your second is in not noticing that the denominators will be different for $(n+1)^{\underline k}$ and $(n+1)^{\underline {k-1}}$: the first has an extra factor of $n+1-k$.

Can you now correct this line: $\Delta_n (n^{\underline k}) = (n+1)^{\underline k} - n^{\underline k} = \frac{(n+1)!}{k!} - \frac{n!}{k!} = \frac{(n+1)!-n!}{k!}$?

Answer 2

You seem to be confused from the start: $n^{\underline k} = n(n-1)\dotsm (n-k+1)$ is not equal to $\frac{n!}{k!}$, but rather to $\frac{n!}{(n-k)!}$.

Looking at the difference $$(n+1)n(n-1)\dotsm (n-k+2) - n(n-1)\dotsm (n-k+2)(n-k+1)$$ is still valid, but there's no denominator we're "forgetting" here; this simply is the difference $(n+1)^{\underline k} - n^{\underline k}$.

You are right that this "is a situation of $ab - bc$", but the common factor $b$ is not all of $n!$, but rather the product $n(n-1)\dotsm (n-k+2) = n^{\underline{k-1}}$. You are right that $a-c = (n+1)-(n-k+1) = k$, so we get $(a-c)b = k n^{\underline{k-1}}$ asour final answer.