Binomial proof of $n! = n^r - n(n - 1)^r + \frac{n(n-1)}{2!}\left(n -2 \right)^r - \dots$ when $r = n$

Question

I realize proofs for this identity have been asked before, but the ones I've found so far involve either induction, generating functions, counting number of functions, etc. I've yet to find this precise argument which I'm struggling a bit with. At any rate, apologies if this counts as a duplicate post.

Again, the claims is,

$$n! = n^r - n(n - 1)^r + \frac{n(n-1)}{2!}\left(n -2 \right)^r - \dots \ (\text{when} \ r = n) $$

Here is the proof:

We have,

$$ (e^x - 1)^n = \left( x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \right)^n \tag{1} $$

Also, by the binomial theorem we have,

$$ (e^x - 1)^n =e^{nx} - ne^{(n-1)x} + \frac{n(n-1)}{2!}e^{(n-2)x} - \dots \tag{2}$$

Expanding each of the terms $e^{nx}, e^{(n-1)x}, \dots $ we see that the coefficient of $x^r$ in (2) is

$$ \frac{n^r}{r!} - n\frac{(n-1)^r}{r!} + \frac{n(n-1)}{2!}\left( \frac{(n - 2)^r}{r!} \right) - \frac{n(n-1)(n-2)}{3!} \left( \frac{(n-3)^r}{r!} \right) + \dots \tag{3}$$

Now equate coefficients of $x^r$ in (1) and (2) and the result follows.

\begin{align} \\ \end{align}

The logic between $(2)$ and $(3)$ isn't quite clicking. "[E]xpanding each of the terms" to me would mean something like $e^{rx} = 1 + \frac{(rx)^2}{2!} + \dots $ for the $r^{\text{th}}$ term. It's unclear to me if we're supposed to expand every binomial term in $(2)$ into a series and then add them all up or how that would produce an $x^r$ term if we did.

If someone could slap some training wheels on the derivation of $(3)$ I'd be much obliged.

Answer 1

Your intuition looks right. I hope this provides clarity to you.

So the second step is done by using binomial theorem on $(e^x-1)^n$. Remember that binomial theorem states $(a+b)^n=\sum_{k=0}^n \binom{n}{k}a^kb^{n-k}$. So when we use this on $(-1+e^x)^n$ we get that this is equivalent to $$\sum_{k=0}^n \binom{n}{k}(-1)^ke^{(n-k)x}$$ $$=e^{nx}-ne^{(n-1)x}+\binom{n}{2}e^{(n-2)x}-\cdots$$

The third step is done by now using the Taylor expansion of $e^x$ as you suggestion. In particular, recall that we know $e^{tx}=\sum_{k=0}^\infty \frac{(tx)^k}{k!}$. So now we can use this fact in our expression from $(2)$. We get that it is equal to $$\sum_{k=0}^n \binom{n}{k}(-1)^k\sum_{j=0}^\infty \frac{((n-k)x)^j}{j!}$$ $$=\sum_{k=0}^n \binom{n}{k}(-1)^k\sum_{j=0}^\infty \frac{(n-k)^jx^j}{j!}$$ $$=\sum_{k=0}^n \sum_{j=0}^\infty \binom{n}{k}(-1)^k\frac{(n-k)^jx^j}{j!}$$ $$=\sum_{j=0}^\infty \sum_{k=0}^n \binom{n}{k}(-1)^k\frac{(n-k)^jx^j}{j!}$$ $$=\sum_{j=0}^\infty x^j\left(\sum_{k=0}^n \binom{n}{k}(-1)^k\frac{(n-k)^j}{j!}\right)$$ So now we can see that the coefficient of $x^n$ is specifically just $\left(\sum_{k=0}^n \binom{n}{k}(-1)^k\frac{(n-k)^n}{n!}\right)$

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To finish up, we can go back to our original expression which was $(e^x-1)^n=\left(x+\frac{x^2}{2!}+\cdots\right)^n=x^n\left(1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots\right)^n$. If we were to expand that big messy power it would look like $x^n\left(1+x(\text{some possibly complicated expressions})\right)$. Luckily the coefficient of $x^n$ is just $1$. Hence, we can just equate the coefficients of $x^n$ to see that $$1=\sum_{k=0}^n \binom{n}{k}(-1)^k\frac{(n-k)^n}{n!}$$ $$n!=\sum_{k=0}^n \binom{n}{k}(-1)^k(n-k)^n$$