I would like to receive some help about the next problem.
The problem:
I'm trying to prove the next equation: $$\sum_{k = 0}^{n} \frac{(-1)^{-k}}{k!(n - k)!} = 0 \quad, n = 1, 2, ...$$
My work until now: $$\sum_{k = 0}^{n} \frac{(-1)^{-k}}{k!(n - k)!} = \sum_{k = 0}^{n} \frac{1}{(-1)^{k}k!(n - k)!} =$$ $$= \frac{1}{(-1)^0 \cdot n!} + \frac{1}{(-1)^1 \cdot (n - 1)!} + \frac{1}{(-1)^2 \cdot 2!(n - 2)!} + \frac{1}{(-1)^3 \cdot 3!(n - 3)!} + \cdot \cdot \cdot + \frac{1}{(-1)^{n - 3} \cdot (n - 3)!3!} + \frac{1}{(-1)^{n - 2} \cdot (n - 2)!2!} + \frac{1}{(-1)^{n - 1} \cdot (n - 1)!} + \frac{1}{(-1)^n \cdot n!} \quad (1)$$
Because we have $n + 1$ summands i tried to see what happens when $n$ is even or odd number:
1) $n = 2k - 1$, $k \in \mathbb{N} \Longrightarrow$ Now there is $2k$ summands
$$(1) \iff \frac{1}{(2k - 1)!}\left( \frac{1}{(-1)^0} + \frac{1}{(-1)^{2k - 1}}\right) + \frac{1}{(2k - 2)!}\left( \frac{1}{(-1)^1} + \frac{1}{(-1)^{2k - 2}}\right) + \frac{1}{2!(2k - 3)!}\left( \frac{1}{(-1)^2} + \frac{1}{(-1)^{2k - 3}}\right) + \cdot \cdot \cdot + \frac{1}{(k - 1)!k!}\left( \frac{1}{(-1)^{k - 1}} + \frac{1}{(-1)^k}\right) = 0$$ This is equal $0$, because the difference in powers of number $-1$ is odd number, so one of them has to be $1$ and another has to be $-1$.
2) $n = 2k$, $k \in \mathbb{N} \Longrightarrow$ Now there is $2k + 1$ summands
$$(1) \iff \frac{1}{(2k)!}\left( \frac{1}{(-1)^0} + \frac{1}{(-1)^{2k}}\right) + \frac{1}{(2k - 1)!}\left( \frac{1}{(-1)^1} + \frac{1}{(-1)^{2k - 1}}\right) + \frac{1}{2!(2k - 2)!}\left( \frac{1}{(-1)^2} + \frac{1}{(-1)^{2k - 2}}\right) + \cdot \cdot \cdot + \frac{1}{(k - 1)!(k + 1)!}\left( \frac{1}{(-1)^{k - 1}} + \frac{1}{(-1)^{k + 1}}\right) + \frac{1}{(-1)^k \cdot k!k!} \quad \Longrightarrow \quad ?$$
Here, i can't use the same princip as in the first case. Also, the last summand is the $k + 1$ -st summand, he is in the middle of this sum and doesn't have a pair.
My question:
Please, could you help me with some advice or a hint that i could use to finnish my proof of the given equation?
Thank you, for your time and your help!
HINT
Actually you were thinking a bit to complicated. Therefore consider the series expansion of $(1-x)^n$ which is due the Binomial Theorem given by
$$\sum_{k=0}^n\binom nk (-x)^k(1)^{n-k}=\sum_{k=0}^n\binom nk (-x)^k$$
Further notice that the binomial coefficient can be written as
$$\binom nk = \frac{n!}{k!(n-k)!}$$
Putting these two together yields to
$$\sum_{k=0}^n\frac{n!}{k!(n-k)!}(-x)^k=n!\sum_{k=0}^n\frac{(-x)^k}{k!(n-k)!}$$
For $x=1$ this equals your given formula. Now look at the original term we have expanded. Can you finish it from hereon?