How to prove the identity: $\sum\limits_{k=0}^{n}(-1)^k\binom{j}{k}=(-1)^n\binom{j-1}{n}$? I'm trying to simplify the following summation by Pascal's identity then got:
$$\sum\limits_{k=0}^{n}(-1)^k\binom{j}{k}=(-1)^0\binom{j-1}{-1}+(-1)^{n}\binom{j-1}{n},$$
that is: all the terms between the two ends (not included) are canceled out, my idea is $\binom{j-1}{-1}$ is undefined and my simplification is wrong. So what's the correct way to prove it?
I also want to know a problem related to this (or should I post another question?): when $j=0$, what should be the value of $\dbinom{j}{k}, 0\le k$?
 A: Another way using generating functions. The identity can be obtained from taking the coefficient of $x^n$ in the expansions of both sides of the identity
$$
\frac{1}{1-x}\times(1-x)^j=(1-x)^{j-1}.
$$
The binomial theorem implies that $[x^n]((1-x)^{j-1})=(-1)^n\binom{j-1}{n}$ where $[x^n]$ means extract the coefficient of $x^n$ in the series.
Since $(1-x)^{-1}=\sum_{m=0}^\infty x^m$ and $(1-x)^j=\sum_{u=0}^j (-1)^u\binom{j}{u}$, the cauchy product implies that $[x^n](\frac{1}{1-x}\times(1-x)^j)=\sum_{k=0}^n(-1)^k\binom{j}{k}$ from which the result follows.
A: You have to take out the case $k=0$(because in Pascal's recursion, this is the base case) so
$$\sum _{k=0}^n(-1)^k\binom{j}{k}=1+\sum _{k=1}^n(-1)^k\binom{j}{k}=1+\sum _{k=1}^n(-1)^k\left (\binom{j-1}{k}+\binom{j-1}{k-1}\right )$$ Now you put together the ones you want to subtract as
$$1-1-\binom{j-1}{1}+\binom{j-1}{1}+\binom{j-1}{2}-\binom{j-1}{2}\cdots +(-1)^{n-1}\binom{j-1}{n-1}+(-1)^{n}\binom{j-1}{n-1}\color{blue}{+(-1)^n\binom{j-1}{n}},$$
where the only one that survives is the $\color{blue}{\text{blue}}$ term.
Notice that $\binom{0}{0}=1$ and $\binom{0}{k}=0$ for $k\neq 0.$

Moved from comments:

It might be helpful to think of binomial as $\frac{n^\underline k}{k!}$, where the numerator is the falling factorial. – Phicar 29 mins ago


$n^\underline k=n(n−1)⋯(n−k+1)$ so if $n=0$ the first terms is $0$. But, if $k=0$, then there are no terms, so you are not multiplying anything. So it is $1$.

A: Simple induction on $n$ will do. The inductive hypothesis is
\begin{eqnarray*}
\sum_{k=0}^{n}  (-1)^{k} \binom{j}{k} = (-1)^n \binom{j-1}{n}.
\end{eqnarray*}
So
\begin{eqnarray*}
\sum_{k=0}^{n}  (-1)^{k} \binom{j}{k} +(-1)^{n+1} \binom{j}{n+1}= (-1)^{n+1} \left( \binom{j}{n+1} -\binom{j-1}{n} \right) =(-1)^{n+1} \binom{j-1}{n+1}.
\end{eqnarray*}
A: The is no one "correct way" to prove this. Here are a couple of related approaches.

Vandermonde's Identity
Using $(-1)^k[k\le n]=(-1)^n\binom{-1}{n-k}$, we get
$$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{j}{k}
&=(-1)^n\sum_{k=0}^n\binom{-1}{n-k}\binom{j}{k}\\[3pt]
&=(-1)^n\binom{j-1}{n}
\end{align}
$$

Cauchy Product
Using the "coefficient of" operator, $\left[x^n\right]$, we get
$$
\begin{align}
\binom{j-1}{n}
&=\left[x^n\right](1+x)^{j-1}\tag1\\[6pt]
&=\left[x^n\right](1+x)^{-1}(1+x)^j\tag2\\[6pt]
&=\sum_{k=0}^n\color{#C00}{\left[x^{n-k}\right](1+x)^{-1}}\color{#090}{\left[x^k\right](1+x)^j}\tag3\\
&=\sum_{k=0}^n\color{#C00}{\binom{-1}{n-k}}\color{#090}{\binom{j}{k}}\tag4\\
&=\sum_{k=0}^n(-1)^{n-k}\binom{j}{k}\tag5
\end{align}
$$
Explanation:
$(1)$: Binomial Theorem
$(2)$: property of exponents
$(3)$: Cauchy Product
$(4)$: Binomial Theorem
$(5)$: $\binom{-1}{n-k}=(-1)^{n-k}[k\le n]$
This approach can be expanded to a proof of Vandermonde's Identity.
A: For completeness, here’s a combinatorial proof. Note that it’s really very similar to the computational proof using Pascal’s identity.
Let $\mathscr{A}=\{A\subseteq[j]:j\in A\}$, where as usual $[j]=\{1,2,\ldots,j\}$; the map
$$\varphi:\mathscr{A}\to\wp([j-1]):A\mapsto A\setminus\{j\}$$
is a bijection and has the property that $|A|$ is odd iff $|\varphi(A)|$ is even (and of course vice versa). Let $\mathscr{S}=\{A\subseteq[j]:|A|\le n\}$, $\mathscr{S}_0=\{A\in\mathscr{S}:|A|\text{ is even}\}$, and $\mathscr{S}_1=\{A\in\mathscr{S}:|A|\text{ is odd}\}$; then
$$\sum_{k=0}^n(-1)^k\binom{j}k=|\mathscr{S}_0|-|\mathscr{S}_1|\,.\tag{1}$$
For each $A\in\mathscr{A}$, one of the sets $A$ and $\varphi(A)$ is in $\mathscr{S}_0$ and the other in $\mathscr{S}_1$, so the pair contributes a net $0$ to $(1)$. This accounts for every member of $\mathscr{S}$ except the subsets of $[j-1]$ of cardinality $n$, which are not in $\varphi[\mathscr{A}\cap\mathscr{S}]$. They are in $\mathscr{S}_0$ if $n$ is even and in $\mathscr{S}_1$ otherwise, and there are $\binom{j-1}n$ of them, so
$$\sum_{k=0}^n(-1)^k\binom{j}k=(-1)^n\binom{j-1}n\,.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\sum_{k = 0}^{n}\pars{-1}^{k}{j \choose k}} =
\sum_{k = 0}^{n}\pars{-1}^{k}{j \choose j - k} =
\sum_{k = 0}^{n}\pars{-1}^{k}\bracks{z^{j - k}}\pars{1 + z}^{j}
\\[5mm] = &\
\bracks{z^{j}}\pars{1 + z}^{j}\sum_{k = 0}^{n}\pars{-z}^{k} =
\bracks{z^{j}}\pars{1 + z}^{j}\,{\pars{-1}^{n + 1}z^{n + 1} - 1 \over
\pars{-z} - 1}
\\[5mm] = &\
\underbrace{\bracks{z^{j}}\pars{1 + z}^{j - 1}}_{\ds{=\ 0}}\ +\
\pars{-1}^{n}\bracks{z^{j - n - 1}}\pars{1 + z}^{j - 1} =
\pars{-1}^{n}{j - 1 \choose j - n - 1}
\\[5mm] = &\
\bbx{\pars{-1}^{n}{j - 1 \choose n}} \\ &
\end{align}
