While studying Stirling numbers of the second kind the following formula for $n!$ suddenly popped up!

$$n! = (-1)^n \sum\_{i=1}^n (-1)^i\binom{n}{i} i^n $$

I was curious if I could find a "direct" proof, i.e. not involving Stirling numbers, but after trying different approaches, f.i. induction, I was stuck. So a hint would be welcome!
['https://math.stackexchange.com/questions/4343329', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/392000/']
Here is some insight for you.

Consider a partially ordered set $S$ which has all meets. We will prove that $S$ has all joins.

Let us consider some indexed collection $\{s\_i \in S\}\_{i \in I}$. Let $J = \{x \in S \mid \forall i \in I (s\_i \leq x)\}$. I claim that $\bigwedge J$ is the join of $s$.

Indeed, we first note that for all $i$, for all $x \in J$, $s\_i \leq x$ (by the very definition of $J$). Therefore, $s\_i \leq \bigwedge J$.

Now suppose that for all $i \in I$, $s\_i \leq x$. Then $x \in J$, so $\bigwedge J \leq x$. This completes the proof. $\square$

Your definition also works. Let's work through it.

First, we must show that $Q := \{ \bigvee X \mid X \subseteq \bigcup\limits\_{\alpha \in \mathcal{J}} I\_\alpha$ and $\bigvee X$ exists$\}$ is in fact a $c$-ideal.

As you have shown, (1) is trivial.

For (2), suppose that $X \subseteq \bigcup\limits\_{\alpha \in \mathcal{J}} I\_\alpha$ and $\bigvee X$ exists, and that $a \leq \bigvee X$. Then $a = a \land \bigvee X = \bigvee \{a \land x \mid x \in X\}$ (using the fact that $a \land$ is the left adjoint to $a \implies$ and therefore preserves any colimits that do exist). Since each $I\_\alpha$ is downward closed, if $x \in I\_\alpha$ then $a \land x \in I\_\alpha$. And thus, we see that $\{a \land x \mid x \in X\} \subseteq \bigcup\limits\_{\alpha \in \mathcal{J}} I\_\alpha$. Thus, $a \in Q$.

For (3), we need to get a big cleverer if we wish to avoid choice. Given $a \in Q$, let $f(a) = \{w \in \bigcup\limits\_{\alpha \in \mathcal{J}} I\_\alpha \mid w \leq a\}$; then clearly, $a = \bigvee f(a)$.

Now suppose we have some family $\{x\_k \in Q\}\_{k \in K}$, and that $\bigvee\limits\_{k \in K} x\_k$ exists. Then in particular, we have $\bigvee\limits\_{k \in K} x\_k = \bigvee\limits\_{k \in K} \bigvee f(x\_k) = \bigvee \bigcup \limits\_{k \in K} f(x\_k)$. Note that $f(x\_k) \subseteq \bigcup\limits\_{\alpha \in \mathcal{J}} I\_\alpha$ by definition, so $\bigcup \limits\_{k \in K} f(x\_k) \subseteq \bigcup\limits\_{\alpha \in \mathcal{J}} I\_\alpha$. Therefore, $\bigvee\limits\_{k \in K} x\_k = \bigvee \bigcup \limits\_{k \in K} f(x\_k) \in Q$.

Now, let's go about proving that $Q$ is in fact the join of the $I\_\alpha$.

Clearly, we see that if $x \in I\_\alpha$, then $x = \bigvee \{x\} \in Q$. So $I\_\alpha \subseteq Q$ for all $\alpha$.

Now suppose that we had some $y$ such that for all $\alpha$, $I\_\alpha \subseteq y$. Then condition 3 ensures that $Q \subseteq y$.
I found an alternative way to define the join. I would still like insight on the original question and/or a relationship to what follows.

Let $A = \{ K \in H' \ | \ \underset{\alpha \in \mathcal{J}}{\bigcup} I\_{\alpha} \subseteq K \}$ that is $A$ is the set of complete ideals containing the union. Now clearly, $A$ is inhabited since $\underset{\alpha \in \mathcal{J}}{\bigcup} I\_{\alpha} \subseteq H \in H'$. Trivially we have the $\bigcap A$ is a complete ideal as it is an intersection of complete ideals. We wish to show that $\underset{\alpha \in \mathcal{J}}{\bigvee} I\_{\alpha} = \bigcap A$. First note that for every $\beta \in \mathcal{J}$ we have $I\_{\beta} \subseteq \underset{\alpha \in \mathcal{J}}{\bigcup} I\_{\alpha} \subseteq \bigcap A$ (where the last inclusion comes from the fact that $\underset{\alpha \in \mathcal{J}}{\bigcup} I\_{\alpha} \subseteq K$ for all $K \in A$.) So clearly, $\bigcap A$ is an upperbound. Suppose you have another upperbound $M$ in $H'$. That is for all $\beta \in \mathcal{J}$ we have $I\_{\beta} \subseteq M$. Then $\underset{\alpha \in \mathcal{J}}{\bigcup} I\_{\alpha} \subseteq M$ and thus $M \in A$. So we can conclude that $\bigcap A \subseteq M$.