Taylor expansion of an stirling identity I have been searching many ways for a week just to solve this, to no avail.
I'm still confused about how the Taylor expansion is produced.
It is so advanced compared to the subjects that I took.
I am currently taking advance researches or work/journals from other mathematicians but I still cannot do this:
$$\frac{(e^{w}-1)^{k}}{k!} = \sum_{n=k}^{\infty }{ n \brace k}
\frac{w^{n}}{n!}.$$
Let us recall that the Stirling numbers satisfy the identities:
$$\begin{array}{rcl}
\displaystyle{ n \brace k} &=& \displaystyle \frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j}j^n = \frac{1}{k!}\sum_{j=0}^{k}(-1)^{j}{k \choose j}(k-j)^{n}
\\\displaystyle{ n+1 \brace k} &=& \displaystyle k{ n \brace
k}+ { n \brace k-1}
\end{array}$$
and appear in the Taylor expansion:
$$\frac{(e^{w}-1)^{k}}{k!} = \sum_{n=k}^{\infty }{ n \brace k}\frac{w^{n}}{n!}.$$
 A: In the following we use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g.
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n\qquad\text{and}\qquad  k^n=n![z^n]e^{kz}
\end{align*}

We obtain
  \begin{align*}
\displaystyle{ n \brace k}&=\frac{1}{k!}\sum_{j=0}^k(-1)^{j-k}\binom{k}{j}j^n\\
&=\frac{1}{k!}\sum_{j=0}^k\binom{k}{j}(-1)^{j-k}n![z^n]e^{jz}\tag{1}\\
&=\frac{n!}{k!}[z^n]\sum_{j=0}^k\binom{k}{j}\left(e^z\right)^j(-1)^{k-j}\tag{2}\\
&=\frac{n!}{k!}[z^n](e^z-1)^k\tag{3}\\
&=n![z^n]\frac{(e^z-1)^k}{k!}
\end{align*}
  and the claim $$\frac{(e^{z}-1)^{k}}{k!} = \sum_{n=k}^{\infty }{ n \brace k}
\frac{z^{n}}{n!}$$ follows.

Comment:


*

*In (1) we apply the coefficient of operator.

*In (2) we do some rearrangements and use the linearity of the coefficient of operator.

*In (3) we apply the binomial theorem.
A: Here is the proof of the EGF from the recurrence. We have
$${n+1\brace k} = k {n\brace k} + {n\brace k-1}.$$
with ${0\brace k} = 0,$ ${n\brace 0} = 0$ and ${0\brace 0} = 1.$
Introduce the mixed generating function
$$G(z, u) = 
1 + \sum_{n\ge 1} \sum_{k\ge 1} {n\brace k} u^k \frac{z^n}{n!}.$$
Multiply the recurrence by $u^k \frac{z^n}{n!}$ and sum over
$n\ge 0, k\ge 1$ to get
$$\sum_{n\ge 0} \sum_{k\ge 1} u^k \frac{z^n}{n!}
(n+1)! [z^{n+1}] [u^k] G(z, u)
\\ = \sum_{n\ge 0} \sum_{k\ge 1} k u^k \frac{z^n}{n!}
n! [z^n] [u^k] G(z, u) +
\sum_{n\ge 0} \sum_{k\ge 1} u^k \frac{z^n}{n!}
n! [z^n] [u^{k-1}] G(z, u).$$
This yields
$$ \sum_{n\ge 0} (n+1) z^n [z^{n+1}]
\sum_{k\ge 1} u^k [u^k] G(z, u)
\\ = \sum_{n\ge 0} z^n [z^n]
\sum_{k\ge 1} k u^k [u^k] G(z, u) +
\sum_{n\ge 0} z^n [z^n]
\sum_{k\ge 1} u^k [u^{k-1}] G(z, u).$$
Continuing we have
$$ \sum_{n\ge 0} (n+1) z^n [z^{n+1}]
(G(z, u) - 1)
\\ = u \sum_{n\ge 0} z^n [z^n]
\sum_{k\ge 1} k u^{k-1} [u^k] G(z, u) +
u \sum_{n\ge 0} z^n [z^n]
\sum_{k\ge 1} u^{k-1} [u^{k-1}] G(z, u)$$
which finally yields
$$\frac{\partial}{\partial z} G(z, u)
= u \sum_{n\ge 0} z^n [z^n] \frac{\partial}{\partial u} G(z, u)
+ u \sum_{n\ge 0} z^n [z^n] G(z, u)$$
or
$$\bbox[5px,border:2px solid #00A000]
{\frac{\partial}{\partial z} G(z, u)
= u \frac{\partial}{\partial u} G(z, u)
+ u G(z, u).}$$
Now introduce $G(z, u) = \exp H_1(z, u)$ to get
$$\exp H_1(z, u) \frac{\partial}{\partial z} H_1(z, u)
= u \exp H_1(z, u) \frac{\partial}{\partial u} H_1(z, u)
+ u \exp H_1(z, u)$$
or
$$\frac{\partial}{\partial z} H_1(z, u)
= u \frac{\partial}{\partial u} H_1(z, u)
+ u.$$
Putting $H_1(z, u) = H_2(z, u) - u$ we obtain
$$\frac{\partial}{\partial z} H_2(z, u)
= u \frac{\partial}{\partial u} H_2(z, u).$$
In the last step we put $H_2(z, u) = u \exp H_3(z)$ to get
$$u \exp H_3(z) \frac{d}{dz} H_3(z)
= u \exp H_3(z)$$
or $$\frac{d}{dz} H_3(z) = 1$$
and $H_3(z) = z + C.$
This yields for $G(z, u)$ the form
$$G(z, u) = \exp(u\exp(z+C)-u).$$
Now we have ${n\brace 1} = 1$ for all $n\ge 1$ which says
$$1 = n! [z^n] [u^1] \exp(u(\exp(z+C)-1))
\\ = n! [z^n] (\exp(z+C)-1)
= \exp(C) n! [z^n] \exp(z) = \exp(C).$$
Therefore $C=0$ and
$$\bbox[5px,border:2px solid #00A000]
{G(z, u) = \exp(u(\exp(z)-1)).}$$
Addendum.  Apparently   the  above  manipulations   to  solve  the
differential  equation are  not quite  rigorous and  the  most general
solution (which is easily verified) is given by
$$\exp(-u) F(u \exp(z))$$
with $F$ an arbitrary function. We now show how to determine $F$.
Let $$F(w) = \sum_{q\ge 0} f_q \frac{w^q}{q!}.$$
Then we have for all $n\ge 0$
$$1 = {n\brace n} = n! [z^n]
[u^n] \exp(-u) F(u \exp(z)).$$
We obtain
$$1 = n! [z^n] \sum_{k=0}^n \frac{(-1)^{n-k}}{(n-k)!}
\frac{1}{k!} f_k \exp(kz)
\\ = \frac{1}{n!} \sum_{k=0}^n {n\choose k} (-1)^{n-k} f_k k^n.$$
We  claim  $f_q=1$   for  all  $q$  with  the   values  determined  by
interpreting the above as a recurrence. The base case for $n=0$ says
$$1 = 1 \times 1 \times (-1)^0 \times f_0 \times 0^0 = f_0$$
and it holds. 
For the induction step we get from the induction hypothesis that
$$1 = \frac{1}{n!} {n\choose n} (-1)^{n-n} n^n f_n +
\frac{1}{n!} \sum_{k=0}^{n-1} {n\choose k} (-1)^{n-k} k^n
\\ = \frac{1}{n!} n^n (f_n - 1) + 
\frac{1}{n!} \sum_{k=0}^{n} {n\choose k} (-1)^{n-k} k^n.$$
Note however that
$$\frac{1}{n!} \sum_{k=0}^{n} {n\choose k} (-1)^{n-k} k^n
= \frac{1}{n!} \sum_{k=0}^{n} {n\choose k} (-1)^{n-k} 
n! [v^n] \exp(kv)
\\ = [v^n] \sum_{k=0}^{n} {n\choose k} (-1)^{n-k} \exp(kv)
= [v^n] (\exp(v)-1)^n = 1$$
because $\exp(v)-1$ starts at $[v^1].$ This finally yields
$$1 = \frac{1}{n!} n^n (f_n - 1) + 1
\quad\text{or}\quad
0 = \frac{1}{n!} n^n (f_n - 1)$$ and hence $f_n=1$ as claimed.
With $F(w)$ shown to be $\exp(w)$ we obtain once more
$$\bbox[5px,border:2px solid #00A000]
{G(z, u) = \exp(u(\exp(z)-1)).}$$
We have used $0^0={0\choose 0}=1$ throughout.
