Discrete Math Identity Proof Binomial Coefficients The question is to prove this identity:
!
where k, m, n ∈ Z+.
Using pascal's identity on the left, so far I have:

!
If m is even then they cancel each other and should equal 0. If m is odd then answer would be (n choose m)
I'm stuck. On the right side, m odd/even doesn't matter neither results in equaling 0. What should I be doing next/or differently?
 A: Let's do this by induction on $m$, where $n$ is fixed.  We see right away that for $m = 0$, we have $$ (-1)^0 \binom{n}{0} = 1 = (-1)^0 \binom{n-1}{0}. $$
Now assume the identity holds for all values $\leq m$, and we will show that it for $m + 1$. As a heads up, we will make use of the binomial recursive formula, sometimes referred to as Pascal's Triangle: $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n - 1}{k}$.
\begin{align*}
\sum\limits_{k = 0}^{m+1}(-1)^k \binom{n}{k} &= (-1)^{m+1}\binom{n}{m+1} + \sum\limits_{k = 0}^{m}(-1)^k \binom{n}{k} \\
&= (-1)^m \binom{n-1}{m} + (-1)^{m+1} \binom{n}{m+1} \text{ by the inductive hypthosis}\\
&= (-1)^{m+1} \left( \binom{n}{m+1} - \binom{n-1}{m} \right) \\
&= (-1)^{m+1} \binom{n-1}{m+1} \text{ by Pascal's Triangle}.
\end{align*}
This completes the proof.

EDIT: A fair amount of identities with Binomial Coefficients can be proven using induction and Pascal's Triangle; if you come across a similar identity, induction and Pascal's Triangle are often a good place to start.
A: Here is a direct proof as well. By Binomial theorem we have $$(1-x)^{n-1}=\sum_{k=0}^{n-1} \binom{n-1}{k}(-1)^k x^k.$$ At the same time, using geometric progression, we have
$$(1-x)^{n-1}=\frac{1}{1-x}(1-x)^n=(1+x+x^2+\dots)\sum_{k=0}^{n} \binom{n}{k}(-1)^k x^k.$$
Comparing coefficients at $x^m$ on boths sides, we get
$$
\binom{n-1}{m}(-1)^m=1\cdot \binom{n}{m}(-1)^m+1\cdot \binom{n}{m-1}(-1)^{m-1}+\dots+1\cdot \binom{n}{0}(-1)^{0} 
$$
