Proving these binomial sums

Question

Need help proving with these two:

1) $$\sum_{k=0}^m (-1)^k \binom{n}{k}= (-1)^m \binom{n - 1}{m}$$

2)

$$\sum_{k=0}^n \binom{n}{k} \binom{k}{m}= \binom{n}{m} 2^{n-m}$$

I tried using these two properties below, but got nowhere. $$\sum_{k=0}^n (-1)^k \binom{n}{k} = 0$$

$$ \sum_{k=0}^n \binom{n}{k} = 2^n$$

Answer 1

For the first equality, it can be easily proved by induction on $m$. When $m = 0$, the LHS is $1$ and the RHS is also $1$, thus the equality holds. For $m \geq 1$, we have \begin{align} \sum_{k=0}^m (-1)^k \binom{n}{k} &= \sum_{k=0}^{m-1}(-1)^k\binom{n}{k} + (-1)^m\binom{n}{m} \\ &= (-1)^{m-1}\binom{n-1}{m-1} + (-1)^{m}\binom{n}{m}\\ &= (-1)^m\left(\binom{n}{m} - \binom{n-1}{m-1}\right) \\ &= (-1)^m \binom{n-1}{m} \end{align}

For the second equality, we have \begin{align} \sum_{k=0}^n \binom{n}{k}\binom{k}{m} &= \sum_{k=0}^n \binom{n}{m} \binom{n-m}{n-k} \\ &= \binom{n}{m} \sum_{k=0}^n \binom{n-m}{n-k} \\ &= \binom{n}{m} 2^{n-m} \end{align}