Why is $ A_1 x + ... + A_n x^n $ a solution of $ \sum_0^{n} (-1)^n \frac{x^n}{n!} \frac{d^n y}{d x^n} = 0 $? I was playing(/fiddling) around with some maths and I saw this pattern(
where $ A_n $ is a constant.):
$ A_1 x $ is a soultion of:
$$ \frac{y}{x} - \frac{dy}{dx} = 0 $$
$ A_1 x + A_2 x^2 $ is a solution of:
$$ \frac{y}{x} - \frac{dy}{dx} + \frac{x}{2!} \frac{d^2y}{dx^2} =0 $$
$ A_1 x + A_2 x^2 + A_3 x^3 $ is a solution of:
$$ \frac{y}{x} - \frac{dy}{dx} + \frac{x}{2!} \frac{d^2y}{dx^2} - \frac{x^2}{3!} \frac{d^3y}{dx^3} =0 $$
It continues so on. Can someone prove the solution of 
$ A_1 x + A_2 x^2 + A_3 x^3 + ... + A_n x^n $ is:
$$ \sum_{k=0}^{n} (-1)^k \frac{x^{k-1}}{k!} \frac{d^k y}{d x^k} = 0 $$  

For any polynomial $P(x)$ of degree at most $n$, we have $\frac{d^k}{dx^k}P(x) = 0$ for any $k > n$. As a result, for any constant $y$, we have
$$\sum_{k=0}^{n} \frac{(y-x)^k}{k!} \frac{d^k}{d x^k}P(x) 
= \sum_{k=0}^{\infty} \frac{(y-x)^k}{k!} \frac{d^k}{d x^k}P(x) = P(y)$$
The last equality is true because the expression in the middle is nothing but
the Taylor expansion of $P(y)$ with respect to the point $x$.
Substitute $y = 0$, we find any polynomial $P(x)$ with $P(0) = 0$ satisfy the ODE
$$\sum_{k=0}^n \frac{(-x)^k}{k!} \frac{d^k}{dx^k} P(x) = P(0) = 0$$
