A differential equation with a Leibniz's formula I found this exercise and I don't know how to solve the first question.

Solve this differential equation :
$$\sum_{k=0}^{n} \binom{n}{k}y^{(k)}=0$$
Deduce that for all $j<n$ :
$$\sum_{k=0}^{n} \binom{n}{k} \binom{k}{j}(-1)^k=0$$

For the first part, I don't know how to start, it looks like a Leibniz formula for the product of functions $y$ and a function with all its derivatives equal to one .... which is strange.
I also thought about the superposition principle, but I don't know if it's a good idea because it would impliy using differential equations of different orders.
So,if you could give me some tips ^^
Thanks in adavance
 A: $$\sum_{k=0}^{n} \binom{n}{k}y^{(k)}=0\tag{1}$$
Let us denote the initial conditions : $a_k:=y^{(k)}(0)$
Let $Y(s)$ be the Laplace Transform of $y$. The L.T. of (1) is:
$$\sum_{k=0}^{n} \binom{n}{k}(s^kY-\sum_{p=0}^{k-1} a_{k-p-1}s^p)=0\tag{2}$$
Recall: $$\mathcal{L} \{f^{(n)}\} =s^n\mathcal{L} \{f\}-s^{n-1}f(0)-s^{n-2}f'(0)-s^{n-3} f''(0)-\ldots -f^{(n-1)}(0)$$
(2) is equivalent to:
$$(\sum_{k=0}^{n} \binom{n}{k}s^k)Y-\sum_{k=0}^{n}\sum_{p=0}^{k-1}  \binom{n}{k} a_{k-p-1}s^p=0\tag{2}$$
$$(1+s)^n Y-\sum_{p=0}^{n-1}b_ps^p=0\tag{2}$$
for certain coefficients $b_k$, giving :
$$Y=\sum_{p=0}^{n-1}b_p\dfrac{s^p}{(1+s)^n}$$
Taking the inverse LT, we get (see an example below):
$$y=\sum_{p=0}^{n-1}c_p\dfrac{1}{p!}e^{-t}P_p(t) \ \text{for certain polynomials} \ P_p \ \text{ with degree}(P_p)=p$$
Overall, we get, for a certain family of coefficients:
$$y=e^{-t}\sum_{p=0}^{n-1}d_pt^p$$
Example: The inverse Laplace Transform of $\dfrac{s^3}{(s+1)^4}$ is $\dfrac16 e^{-t}(t^3-9t^2+18t-6)$ as can be verified for example with Wolfram Alpha by asking InverseLaplaceTransform[s^3/(s+1)^4,s,t].
