Proof with combinatorial argument 
Show with combinatorial argument that this is equal :
$$\dbinom{n}{k+1} = \dbinom{n-1}{k}+ \dbinom{n-2}{k} +...+ \dbinom{k}{k}$$ 

I have no idea how to do that so it would be really helpful if anyone knows some way.

Suppose we want to distribute $n - k - 1$ identical balls into $k + 2$ distinct boxes. With stars and bars, there are $\binom{n - k - 1 + k + 1}{k + 1} = \binom{n}{k + 1}$ ways to do this.
We can also first distribute $0 \le i \le n - k - 1$ balls into the first box and then distribute remaining the balls into the remaining $k + 1$ boxes. The number of ways to do this is, using stars and bars for the remaining balls and boxes, is
$$\binom{(n - k - 1) + k}{k} + \binom{(n - k - 2) + k}{k} + \dots + \binom{(n - k - n) + k}{k}$$
$$= \binom{n - 1}{k} + \binom{n - 2}{k} + \dots + \binom{k}{k}$$
Since both ways of counting are equivalent, we have
$$\binom{n}{k + 1} = \binom{n - 1}{k} + \binom{n - 2}{k} + \dots + \binom{k}{k}$$
