Alternating sum of binomial coefficients up to certain value I would like to compute the value of the following sum:

$\sum_{k\leq m} (-1)^k\binom{n}{k}$ 

for an integer m and a non negative integer n.
Progress
I know $\sum_{k=0}^n (-1)^k\binom{n}{k} = 0$ and symmetry of binomial-coefficients can be used to simplify some cases but beyond that i'm stuck.  
 A: So $\sum \limits_{k=0}^{m} (-1)^k \binom{n}{k}= \binom{n}{0}-\binom{n}{1}+\binom{n}{2} -\binom{n}{3} + \cdots + (-1)^m \binom{n}{m}$
Decompose every term using $\binom{n}{k} = \binom{n-1}{k}+\binom{n-1}{k-1}$
Which gives you $\binom{n}{0}-\binom{n-1}{0}-\binom{n-1}{1}+\binom{n-1}{1}+\binom{n-1}{2} -\binom{n-1}{2} -\binom{n-1}{3} + \cdots $
By the telescope effect and knowing that $\binom{n}{0} =1$ for all $n$.
We arrive at final answer $(-1)^m \binom{n-1}{m}$ if $m<n$ and $0$ if $m=n$.
A: The answer is $(-1)^m\binom{n-1}{m}$. The trick is to use Pascal's Formula and induction. 
Here's the full argument. For $m=0$, the sum is just equal to $1$ which is indeed the same as $(-1)^0\binom{n-1}{0}$. Now, for the inductive step, assume $m\geq1$ and that the statement is true for smaller values. We get
$$\sum_{k=0}^m (-1)^k\binom{n}{k} = (-1)^m\binom{n}{m} + \sum_{k=0}^{m-1} (-1)^k\binom{n}{k}$$
Now, applying the inductive hypothesis, we get that this is equal to 
$$(-1)^m\binom{n}{m} + (-1)^{m-1}\binom{n-1}{m-1} = (-1)^{m}\left(\binom{n}{m} -
 \binom{n-1}{m-1}\right).$$
By Pascal's Formula, $\binom{n}{m}-\binom{n-1}{m-1}$ is equal to $\binom{n-1}{m}$ and so we get
$$(-1)^m\binom{n-1}{m}$$
and we are done. 
