$ \sum_{l=1}^n(-1)^{l-1}\binom{n}{l}\sum_{k=0}^{l-1}\binom{l-1}{k}b_k = \sum_{k=0}^{n-1}(-1)^k b_k \,. $ Today I encountered the following identity during some work: letting $b_k$ be a sequence,
$$
\sum_{l=1}^n(-1)^{l-1}\binom{n}{l}\sum_{k=0}^{l-1}\binom{l-1}{k}b_k = \sum_{k=0}^{n-1}(-1)^k b_k \,.
$$
I've struggled for a while with properties of binomial coefficients and induction strategies, then I think I found a proof using derivatives (see below). 
Are there more conventional/straightforward proofs?
 A: Actually, there is a way by direct manipulation of the binomials.
$$ \bbox[lightyellow] {  
\eqalign{
  & \sum\limits_{l = 1}^n {\left( { - 1} \right)^{\,l - 1} \left( \matrix{
  n \cr 
  l \cr}  \right)\sum\limits_{k = 0}^{l - 1} {\left( \matrix{
  l - 1 \cr 
  k \cr}  \right)b_{\,k} } }  =   \quad \quad (1) \cr 
  &  = \sum\limits_{1\, \le \,l\,\left( { \le \,n} \right)} {\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,l - 1} \right)} {\left( { - 1} \right)^{\,l - 1} \left( \matrix{
  n \cr 
  l \cr}  \right)\left( \matrix{
  l - 1 \cr 
  k \cr}  \right)b_{\,k} } }  =  \quad \quad (2)  \cr 
  &  = \sum\limits_{0\, \le \,l\,\left( { \le \,n - 1} \right)} {\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,l} \right)} {\left( { - 1} \right)^{\,l} \left( \matrix{
  n \cr 
  l + 1 \cr}  \right)\left( \matrix{
  l \cr 
  k \cr}  \right)b_{\,k} } }  =  \quad \quad (3)  \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,n - 1} \right)} {\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,l} \right)} {\left( { - 1} \right)^{\,l} \left( \matrix{
  n \cr 
  l + 1 \cr}  \right)\left( \matrix{
  l \cr 
  l - k \cr}  \right)b_{\,k} } }  =  \quad \quad (4)  \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,n - 1} \right)} {\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,l} \right)} {\left( { - 1} \right)^{\,l} \left( \matrix{
  n \cr 
  l + 1 \cr}  \right)\left( { - 1} \right)^{\,l - k} \left( \matrix{
   - k - 1 \cr 
  l - k \cr}  \right)b_{\,k} } }  =  \quad \quad (5)  \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,l} \right)} {\left( {\sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,n - 1} \right)} {\left( \matrix{
  n \cr 
  l + 1 \cr}  \right)\left( \matrix{
   - k - 1 \cr 
  l - k \cr}  \right)} } \right)\left( { - 1} \right)^{\,k} b_{\,k} }  =  \quad \quad (6)  \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,l} \right)} {\left( {\sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,n - 1} \right)} {\left( \matrix{
  n \cr 
  n - 1 - l \cr}  \right)\left( \matrix{
   - k - 1 \cr 
  l - k \cr}  \right)} } \right)\left( { - 1} \right)^{\,k} b_{\,k} } \quad \left| {\;0 \le n} \right.\quad  =  \quad \quad (7)  \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n - 1} \right)} {\left( \matrix{
  n - k - 1 \cr 
  n - k - 1 \cr}  \right)\left( { - 1} \right)^{\,k} b_{\,k} } \quad \left| {\;0 \le n} \right.\quad  =   \quad \quad (8) \cr 
  &  = \sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( { - 1} \right)^{\,k} b_{\,k} } \quad \left| {\;1 \le n} \right.  \quad \quad (9) \cr} 
}$$
where:
 (1) original expression;
 (2) we get rid (put in brackets) of the summation bounds
which are implicit in the binomial;
 (3) we change $l$ with $l-1$;
 (4) since $0 \le l$ we can apply symmetry and get rid
also of the lower bound on $l$;
 (5) Upper Negation;
 (6) exchange order of summation;
 (7) symmetry on the first binomial, provided that $0 \le n$;
 (8) bounds allow to apply convolution in $l$;
 (9) the binomial provides the bounds on summation
Concerning the approach you propose, it looks "formally" pretty valid.
I am hesitant from a rigorous point of view when it comes  to take the multiplicative inverse, but I am not a theorist.
However, yours parallels a slight different approach that I would take, based on the Right Shift Operator, 
which I know is valid also from a rigorous point of view.
$$ \bbox[lightyellow] {  
E\,f(x) = f(x + 1)\quad E\;b_{\,k}  = b_{\,k + 1} \quad b_{\,k}  = E^{\,k} \;b_{\,0} 
} $$
Then
$$ \bbox[lightyellow] {  
\eqalign{
  & LHS = \sum\limits_{l = 1}^n {\left( { - 1} \right)^{\,l - 1} \left( \matrix{
  n \cr 
  l \cr}  \right)\sum\limits_{k = 0}^{l - 1} {\left( \matrix{
  l - 1 \cr 
  k \cr}  \right)b_{\,k} } }  = \sum\limits_{l = 1}^n {\left( { - 1} \right)^{\,l - 1} \left( \matrix{
  n \cr 
  l \cr}  \right)\sum\limits_{k = 0}^{l - 1} {\left( \matrix{
  l - 1 \cr 
  k \cr}  \right)E^{\,k} \;b_{\,0} } }  =   \cr 
  &  = b_{\,0} \sum\limits_{l = 1}^n {\left( { - 1} \right)^{\,l - 1} \left( \matrix{
  n \cr 
  l \cr}  \right)\left( {I + E} \right)^{\,l - 1} }  =  - {{b_{\,0} } \over {\left( {I + E} \right)}}\sum\limits_{l = 1}^n {\left( { - 1} \right)^{\,l} \left( \matrix{
  n \cr 
  l \cr}  \right)\left( {I + E} \right)^{\,l} }  =   \cr 
  &  =  - {{b_{\,0} } \over {\left( {I + E} \right)}}\left( {\left( {I - \left( {I + E} \right)} \right)^{\,n}  - I} \right) =  - b_{\,0} {{\left( { - 1} \right)^{\,n} E^{\,n}  - I} \over {\left( {I + E} \right)}} = b_{\,0} {{I - \left( { - 1} \right)^{\,n} E^{\,n} } \over {\left( {I + E} \right)}} \cr} 
} $$
and
$$ \bbox[lightyellow] {  
RHS = \sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( { - 1} \right)^{\,k} b_{\,k} }  = b_{\,0} \sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( { - 1} \right)^{\,k} E^{\,k} }  = b_{\,0} {{I - \left( { - 1} \right)^{\,n} E^{\,n} } \over {I + E}}
} $$
A: We start from
$$\sum_{l=1}^n (-1)^{l-1} {n\choose l}
\sum_{k=0}^{l-1} {l-1\choose k} b_k
= \sum_{k=0}^{n-1} b_k
\sum_{l=k+1}^n {n\choose l} (-1)^{l-1} {l-1\choose k}.$$
We get for the coefficient on $b_k$
$$\sum_{l=k+1}^n {n\choose l} (-1)^{l-1} {l-1\choose k}
= \sum_{l=k}^{n-1} {n\choose l+1} (-1)^l {l\choose k}
\\ = \sum_{l=k}^{n-1} {n\choose l+1} (-1)^l [z^k] (1+z)^l.$$
We may  start $l$ at zero  because $[z^k] (1+z)^l$ is  zero when $0\le
l\lt k$, getting
$$[z^k] \sum_{l=0}^{n-1} {n\choose l+1} (-1)^l (1+z)^l
= - [z^k] \frac{1}{1+z}
\sum_{l=0}^{n-1} {n\choose l+1} (-1)^{l+1} (1+z)^{l+1}
\\ = - [z^k] \frac{1}{1+z}
\left((1-(1+z))^n - 1\right)
= [z^k] \frac{(-1)^{n+1} z^n}{1+z} + [z^k] \frac{1}{1+z}.$$
Note however that $n\gt k$ so there is no contribution from
the first term and we are left with
$$[z^k] \frac{1}{1+z} = (-1)^k$$
as claimed.
A: For a given sequence $b_k$, by Borel's lemma (see here or here), there exists a function $f(x)$ such that
$
b_k = f^{(k)}(0) \equiv \partial^kf.
$
Thus we need only prove
$$
\sum_{l=1}^n(-1)^{l-1}\binom{n}{l}\sum_{k=0}^{l-1}\binom{l-1}{k}\partial^k = \sum_{k=0}^{n-1}(-1)^k \partial^k\,.
$$
The right-hand side can be evaluated formally using $1-x+x^2+\cdots+(-x)^{n-1}=(1-(-x)^n)/(1+x)$, and yields
$$
\frac{1-(-\partial)^n}{1+\partial}\,.
$$
The left-hand side can be manipulated using Newton's binomial formula, giving
$$
\sum_{l=1}^n(-1)^{l-1}\binom{n}{l}(1+\partial)^{l-1}=\frac{1}{-1-\partial}\sum_{l=1}^n\binom{n}{l}(-1-\partial)^l=\frac{(1-1-\partial)^n-1}{-1-\partial}=\frac{1-(-\partial)^n}{1+\partial}\,.
$$
