A closed form of Eulerian numbers

Start from the bivariate generating function of the Eulerian numbers: $$G(z, u) = \frac{u-1}{u-\exp((u-1)z)} = \left(1-\frac{1}{u}\right)\frac{1}{1-\exp((u-1)z)/u}.$$

This is $$\left(1-\frac{1}{u}\right) \sum_{q\ge 0} \frac{\exp(q(u-1)z)}{u^q}.$$

This implies that $$\left\langle {n\atop k}\right\rangle = n! [u^k] \left(1-\frac{1}{u}\right) \sum_{q\ge 0} \frac{q^n(u-1)^n}{n!\times u^q}$$ which is $$[u^k] \frac{(u-1)^{n+1}}{u} \sum_{q\ge 0} \frac{q^n}{u^q} = [u^{k+1}] (u-1)^{n+1} \sum_{q\ge 0} \frac{q^n}{u^q}.$$

Extracting coefficients from this we obtain $$\sum_{p=k+1}^{n+1} {n+1\choose p} (-1)^{n+1-p} (p-(k+1))^n = \sum_{p=k+1}^{n+1} {n+1\choose n+1-p} (-1)^{n+1-p} (p-(k+1))^n.$$

Re-index the sum putting $n+1-p = q$ to get $$\sum_{q=0}^{n-k} {n+1\choose q} (-1)^q (n+1-q-(k+1))^n \\ = \sum_{q=0}^{n-k} {n+1\choose q} (-1)^q (n-k-q)^n$$

We digress for a moment to evaluate $$\sum_{q=0}^{n+1} {n+1\choose q} (-1)^q (n-k-q)^n$$ using the integral $$(n-k-q)^n = \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \exp((n-k-q)z) \; dz.$$ We get for the sum $$\frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \sum_{q=0}^{n+1} {n+1\choose q} (-1)^q \exp((n-k-q)z) \; dz \\ = \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp((n-k)z)}{z^{n+1}} \sum_{q=0}^{n+1} {n+1\choose q} (-1)^q \exp(-qz) \; dz \\ = \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{\exp((n-k)z)}{z^{n+1}} \left(1-\exp(-z)\right)^{n+1} \; dz.$$

Now note that $1-\exp(-z)$ starts at $z$ and therefore $(1-\exp(-z))^{n+1}$ starts at $z^{n+1}$ which means that the integrand is entire, making the integral evaluate to zero, so the sum is zero.

Ending the digression we now observe that the sum from the generating function is in fact $$-\sum_{q=n-k+1}^{n+1} {n+1\choose q} (-1)^q (n-k-q)^n.$$ Re-index one more time to get $$-\sum_{q=0}^k {n+1\choose n+1-q} (-1)^{n+1-q} (n-k-(n+1-q))^n \\ = \sum_{q=0}^k {n+1\choose q} (-1)^{n-q} (q-1-k)^n = \sum_{q=0}^k {n+1\choose q} (-1)^q (k+1-q)^n.$$

This is what was required. QED.

(We count descents instead of ascents. Let $A(n, k)$ denote the number of permutations on $[n]$ with $k-1$ descents)

Here's a combinatorial proof via inclusion-exclusion the essence of which can be found in Miklos Bona's Combinatorics of Permutations. He attributes this proof to Richard Stanley and Hugh Thomas (independently).

Define a "$k$-augmented word on $[n]$" to be an object you get by writing down $k-1$ bars and allocating each element of $[n] = \{1,2, \ldots , n\}$ to one of the $k$ compartments created by the bars and then ordering the set of numbers in each compartment in ascending order. For example, if $n=5$ and $k =4$, then each of the following is a $4$-augmented word on $[5]$: $$ w_{1} = 12||34|5, \qquad w_{2}= |15|4|23, \qquad w_{3}=1|23|5|4 $$

We make two further definitions involving bars:

1) Wall: A bar that is not immediately followed by another bar. (For example, all the bars in the above examples are walls except for the first bar in $w_{1}.$)

2) Extraneous: A bar in a $k$-augmented word is called extraneous if after its deletion, it produces a $(k-1)$-augmented word. (For example, the third bar in the $w_{1}$ is extraneous because the resultant object is a $2$-augmented word. Similarly, so is the first bar in $w_{2}$. However, the second bar in $w_{2}$ is not extraneous.)

In this above vocabulary, it can be clearly seen that the permutations on $[n]$ with $k-1$ descents are in bijection with the $k$-augmented words on $[n]$ that have no extraneous walls. (Each such permutation gives rise to a $k$-augmented word by placing a bar at each of the descent indices. Conversely, the deletion of all the bars in a $k$-augmented word gives rise to a permutation with $k-1$ descents.)

We are now ready to use the principle of inclusion-exclusion on $X$, the set of all possible $k$-augmented words. Note that since each of the $n$ numbers have one of $k$ compartments to live in, then $|X| = k^{n}.$ Below, we use "position" to mean one of the $n+1$ gaps in between a permutation written as a word.

Let $w \in X$ have property $P_{i}$ (for a given $i \in \left\{0,1,\ldots,n\right\}$) if there is an extraneous wall after the $i^{\text{th}}$ symbol of the word. Note that not more than $k-1$ properties can hold simultaneously. Let $S = \{i_{1}< \ldots < i_{j}\}$ be a subset of $\left\{0,1,\ldots,n\right\}$ with $j \leq k-1$, and let $A_S= \left|\{w \in X \text{ satisfying } P_{i_{1}}, \ldots , P_{i_{j}}\}\right|$; then, we claim that $|A_{S}| = (k-j)^{n}$. Indeed, it suffices to find a bijective correspondence between $A_{S}$ and the set of all $(k-j)$-augmented words on $[n]$. If $w \in A_{S}$, then deleting a bar in $w$ at each position belonging to $S$ we get a $(k-j)$-augmented word on $[n]$. Conversely, starting with a $(k-j)$-augmented word on $[n]$, insert a bar at each position belonging to $S$ (and if a bar already exists at that position, insert it to the immediate right of this bar); then, each position of $S$ contains an extraneous wall giving rise to $w \in A_{S}$. Recall that the set of permutations on $n$ with $k-1$ descents equals the number of $k$-augmented words on $[n]$ with no extraneous walls which, by the principle of inclusion-exclusion (using the claim we just showed), equals:

$$A(n, k) = \displaystyle \sum_{j=0}^{k-1}(-1)^{j}\binom{n+1}{j}(k-j)^{n} .$$