proving $ \binom{n}{0}-\binom{n}{1}+\binom{n}{2}+\cdots \cdots +(-1)^{n-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}$ proving $\displaystyle \binom{n}{0}-\binom{n}{1}+\binom{n}{2}+\cdots \cdots +(-1)^{\color{red}{m}-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}.$
$\displaystyle \Rightarrow 1-n+\frac{n(n-1)}{2}+\cdots \cdots (-1)^{n-1}\frac{n.(n-1)\cdot (n-2)\cdots(n-m+2)}{(m-1)!}$
Added
writting LHS as $\displaystyle \binom{n}{0}-\left(\binom{n-1}{0}+\binom{n-1}{1}\right)+\left(\binom{n-1}{1}+\binom{n-1}{2}\right)+\cdots \cdots +(-1)^{n-1}\left(\binom{n-1}{m-2}+\binom{n-1}{m-1}\right)=(-1)^{m-1}\binom{n-1}{m-1}.$
$\displaystyle \binom{n}{0}-\binom{n-1}{0}+\binom{n-1}{1}-\cdots +(-1)^{n-1}\binom{n-1}{m-2}-\left(\binom{n-1}{1}-\binom{n-1}{2}+\cdots +(-1)^n\binom{m-1}{m-1}\right)$ 
wan,t be able to solve after that, help me to solve it
 A: Extended HINT: The result is incorrect as originally stated; it should read
$$\binom{n}0-\binom{n}1+\binom{n}2-\ldots+(-1)^{\color{crimson}m-1}\binom{n}{m-1}=(-1)^{m-1}\binom{n-1}{m-1}\;,$$
or, more compactly,
$$\sum_{k=0}^{m-1}(-1)^k\binom{n}k=(-1)^{m-1}\binom{n-1}{m-1}\;.\tag{1}$$
Fix $n\in\Bbb N$. For $m=1$ the desired result is
$$\binom{n}0=(-1)^0\binom{n-1}0\;,$$
which is indeed true, since both sides are equal to $1$. Suppose as an induction hypothesis that $(1)$ holds for some $m$; for the induction step we want to prove that
$$\sum_{k=0}^m(-1)^k\binom{n}k=(-1)^m\binom{n-1}m\;.\tag{2}$$
Using the induction hypothesis we can rewrite the lefthand side of $(2)$:
$$\sum_{k=0}^m(-1)^k\binom{n}k=(-1)^m\binom{n}m+\sum_{k=0}^{m-1}(-1)^k\binom{n}k=(-1)^m\binom{n}m+(-1)^{m-1}\binom{n-1}{m-1}\;,$$
so to complete the induction step we need only show that
$$(-1)^m\binom{n}m+(-1)^{m-1}\binom{n-1}{m-1}=(-1)^m\binom{n-1}m\;.$$
This is easily done using one of the most basic identities involving binomial coefficients.
Added: After some thought I realize that $(1)$ can be proved by direct calculation:
$$\begin{align*}
\sum_{k=0}^{m-1}(-1)^k\binom{n}k&=\sum_{k=0}^{m-1}(-1)^k\left(\binom{n-1}k+\binom{n-1}{k-1}\right)\\
&=\sum_{k=0}^{m-1}(-1)^k\binom{n-1}k+\sum_{k=0}^{m-1}(-1)^k\binom{n-1}{k-1}\\
&=\sum_{k=0}^{m-1}(-1)^k\binom{n-1}k+\sum_{k=1}^{m-1}(-1)^k\binom{n-1}{k-1}\\
&=\sum_{k=0}^{m-1}(-1)^k\binom{n-1}k+\sum_{k=0}^{m-2}(-1)^{k+1}\binom{n-1}k\\
&=\sum_{k=0}^{m-1}(-1)^k\binom{n-1}k-\sum_{k=0}^{m-2}(-1)^k\binom{n-1}k\\
&=(-1)^{m-1}\binom{n-1}{m-1}+\sum_{k=0}^{m-2}(-1)^k\binom{n-1}k-\sum_{k=0}^{m-2}(-1)^k\binom{n-1}k\\
&=(-1)^{m-1}\binom{n-1}{m-1}\;.
\end{align*}$$
A: HINT: Fix a random $n$ and do induction on $m$.
