TITLE: Is there a combinatorial explanation for the identity: $\sum_{k=1}^n c(n,k)2^k = (n+1)!$
QUESTION [11 upvotes]: Let $c(n,k)$ be the unsigned Stirling numbers of the first kind, i.e., the number of 
$n$-permutations with exactly $k$ cycles.
Apparently,  $$\sum_{k=1}^n c(n,k)2^k = (n+1)!$$
I want to prove the equality. 
I am most interested in a combinatorial explanation.  
The exponential generating function for the RHS is $\frac1{(1-x)^2}$.  Is there a way to derive the e.g.f. for the LHS symbolically?

REPLY [4 votes]: Here is a tedious but extremely elementary combinatorial argument for the more general result.
Let $\pi$ be a permutation of $[n]$ having $k$ cycles. The standard representation of $\pi$ is 
$$(a_{11}a_{12}\ldots a_{1m_1})(a_{21}a_{22}\ldots a_{2m_2})\ldots(a_{k1}a_{k2}\ldots a_{km_k})\;,\tag{1}$$ 
where $a_{i1}>a_{ij}$ for $i=1,\ldots k$ and $j=2\ldots,m_i$, and $a_{11}<a_{21}<\ldots a_{k1}$. In other words, each cycle is listed with its largest element first, and the cycles are listed in increasing order of their largest elements. One-element cycles are not suppressed. The map that sends $\pi$ to the permutation whose one-line representation is $(1)$ without the parentheses is a bijection.
Now we’ll count the permutations $\pi$ with $k$ cycles as in $(1)$. Clearly $a_{k1}=n$, but we have a free choice for $a_{11},a_{21},\ldots,a_{k-1,1}$. Once they’re chosen, in how many ways can we fill out $(1)$? We can ignore the parentheses, and of course we have to put $a_{11}$ in the first slot. We now place the $a_{11}-1$ elements of $[n]$ that are less than $a_{11}$; they can go anywhere in $(1)$ after $a_{11}$ so they can be placed in $(n-1)(n-2)\ldots(n-a_{11}+1)$ different ways. Once they’ve been placed, $a_{21}$ must go in the first free slot. That leaves $n-a_{11}-1$ open slots in $(1)$, and the $a_{21}-a_{11}-1$ elements of $[n]$ that are larger than $a_{11}$ and smaller than $a_{21}$ can go in any of them. Thus, they can be placed in 
$$(n-a_{11}-1)(n-a_{11}-2)\ldots(n-a_{21}+1)$$
different ways, after which $a_{31}$ must be placed in the first available slot.
In general, once we’ve made the forced placement of $a_{i1}$, the $a_{i1}-a_{i-1,1}-1$ elements of $[n]$ lying strictly between $a_{i-1,1}$ and $a_{i1}$ can be placed in
$$(n-a_{i-1,1}-1)(n-a_{i-1,1}-2)\ldots(n-a_{i1}+1)$$
different ways. Thus, if we set $a_{01}=0$, the number of permutations with these values of $a_{11},\ldots,a_{k1}$ is
$$\prod_{i=0}^{k-1}(n-a_{i-1,1}-1)(n-a_{i-1,1}-2)\ldots(n-a_{i1}+1)=\frac{(n-1)!}{\prod_{i=1}^{k-1}(n-a_{k1})}\;.\tag{2}$$ 
As we run over all possible choices for the leading elements $a_{11},\ldots,a_{k1}$, the denominator in $(2)$ runs over all $(k-1)$-fold products of distinct elements of $[n-1]$, so $(2)$ itself runs over all $(n-k)$-fold products of distinct elements of $[n-1]$, and $n\brack k$ is therefore the sum of all such products.
However, that sum is clearly also the coefficient of $x^k$ in the product
$$x^{\overline n}=\prod_{i=0}^{n-1}(x+i)=x(x+1)(x+2)\ldots(x+n-1)\;,$$
so in general we have the identity
$$x^{\overline n}=\sum_{k=0}^n{n\brack k}x^k\;,$$
and the desired identity follows by setting $x=2$.