Bonferroni Inequalities 
Let $k$ and $m$ be positive integers with $k>m$.
  Then the partial sums of
  $$
1-\binom{k}{1} + \binom{k}{2} - \cdots (-1)^m\binom{k}{m}
$$
  has alternating signs.

(The partial sums of the given sum are $P_1=1$, $P_2=1-\binom{k}{1}$, $P_3=1-\binom{k}{1}+\binom{k}{2}$, etc)
I arrived at the above problem while trying to show the follwoing (known as Bonferroni inequalities):
Let $A_1, \ldots, A_n$ be events of a probability space. For a subset $I$ of $\{1,\ldots, n\}$, write $A_I$ to denote $\bigcap_{j\in I}A_j$. Further, denote $\sum_{|I|=i}P(A_I)$ as $\sigma_i$. We agree by convention that $\sigma_0=1$.
Then the partial sums of 
$$
P(A_1^c\cap A_2^c\cap \cdots\cap A_n^c)-\sigma_0+\sigma_1-\sigma_2\cdots
$$
have alternating signs. 
 A: Let $f(k,m)$ be the number of ways to choose $m$ out of $k$ objects, without choosing the last one.
By inclusion-exclusion, $f(k,m)=\binom{k}{m}-f(k,m-1)=(-1)^mP_{m+1}$.
But $f(k,m)$ is obviously $\binom{k-1}{m}$. Therefore $P_{m+1}=(-1)^m\binom{k-1}{m}$.
A: $$\begin{align}
\require{cancel}
&1-\binom{k}{1} + \binom{k}{2} - \cdots (-1)^m\binom{k}{m}\\
&=1+\sum_{r=1}^{m}(-1)^k\binom kr\\
&=1+\sum_{r=1}^{m}(-1)^r\left[\binom {k-1}{r-1}+\binom {k-1}r\right]\\
&=\color{lightgrey}{\cancel{1}-\left[\cancel{{\binom {k-1}0}}+\bcancel{\binom {k-1}1}\right]
+\left[\bcancel{\binom {k-1}1}+\cancel{\binom {k-1}2}\right]
-\left[\cancel{\binom {k-1}2}+\bcancel{\binom {k-1}3}\right]
+\cdots 
+(-1)^m\left[\bcancel{\binom {k-1}{m-1}}+\binom {k-1}m\right]}\\
&=(-1)^m\binom {k-1}m\qquad \blacksquare\\
\end{align}$$

Alternatively:
$$\begin{align}
&1-\binom{k}{1} + \binom{k}{2} - \cdots (-1)^m\binom{k}{m}\\
&=1+\sum_{r=1}^{m}(-1)^k\binom kr\\
&=1+\sum_{r=1}^{m}(-1)^r\left[\binom {k-1}{r-1}+\binom {k-1}r\right]\\
&=1+\sum_{r=0}^{m-1}(-1)^{r+1}\binom {k-1}{r}+\sum_{r=1}^{m}(-1)^r\binom {k-1}r\\
&=1\color{blue}{-\sum_{r=0}^{m-1}(-1)^r\binom {k-1}{r}}
\color{green}{+\sum_{r=1}^{m}(-1)^r\binom {k-1}r}\\
&=\cancel{1}\color{blue}{-\cancel{1}-\bcancel{\sum_{r=1}^{m-1}(-1)^r\binom {k-1}{r}}}
\color{green}{+\bcancel{\sum_{r=1}^{m-1}(-1)^r\binom {k-1}r}+(-1)^m\binom {k-1}m}\\
&=(-1)^m\binom {k-1}m\qquad \blacksquare\\
\end{align}$$
