Equations like Taylor Formula for polynomials 
Consider the linear maps $D: P_n(\mathbb{R}) \to P_n(\mathbb{R})$ given by
$$(Dp)(x) = \sum_{i = 1}^{n}ic_i x^{i-1}$$
and $A: P_n(\mathbb{R}) \to P_n(\mathbb{R})$ given by
$$(Ap)(x) = \sum_{i=0}^{n}c_i (x+1)^i,$$
where $p(x) = \sum_{i=0}^{n}c_i x^i$
Prove that
$$1 + \frac{D}{1!} + \cdots + \frac{D^n}{n!} = A.$$

I think that is a mistake in the question. I cannot see how this equality is true. Seems that $D$ is the derivative operator. Take $p(x) = 1 + x + (x-1)^2$, so
$$(Dp)(x) = 1 + 2(x-1);\quad(D^2p)(x) = 2.$$
Then
$$1 + D + \frac{D^2}{2} = 1 + 1 + 2x - 2 + 1 = 1 + 2x$$
and
$$p(x+1) = 2 + x + x^2.$$
Actually, the degree are differents so, in fact, the equality is not true. I tried to use
$$\sum_{k=0}^{n}\frac{D^k}{k!} = D^0 + \frac{D}{1!} + \cdots + \frac{D^n}{n!} = A.$$
In general,
$$\frac{D^kp}{k!} = \sum_{i=k}^{n}\binom{i}{k}c_i x^{i-k}.$$
Then
$$\sum_{k=1}^{n}\frac{D^k}{k!} = \sum_{k=0}^{n}\sum_{i=k}^{n}\binom{i}{k}c_i x^{i-k}.$$
This is almost
$$A = \sum_{i=0}^{n}c_i \sum_{k=0}^{n}\binom{i}{k}x^{i-k}.$$
Can someone give me a hint? I stuck on this part.
 A: We can prove it by induction: for $n=1$ thesis is $1+D=A$ and seeing the generic polinomial $p(x)=ax+b$ you have $Ap(x)=p(x+1)=ax+a+b$ and $(1+D)(ax+b)=ax+b+Dp(x)=ax+b+a$.
Suppose now thesis true for such $n=k$. Suppose now $n=k+1$. If $p$ has degree $\leq k+1$ we can write $p(x)=ax^{k+1}+q(x)$ with $\deg(q) \leq k$. Remembering that we have $D^{k+1}q\equiv 0$ we can see that: $$\left(\sum_{i=0}^{k+1} \frac {D^i}{i!}\right)p(x)= \left(\sum_{i=0}^{k+1} \frac{D^i}{i!}\right)(ax^{k+1}+q(x))= a\left(\sum_{i=0}^{k+1} \frac{D^i}{i!}\right)x^{k+1}+Aq(x)$$
So we have just to demonstrate that $\left(\sum_{i=0}^{k+1} \frac{D^i}{i!}\right)x^{k+1}=(x+1)^{k+1}$ Can you show this?
A: $D^{0}$ is supposed to be the identity operator in this formula. So $D^{0}p(x)=p(x)=1+x+(x-1)^{2}$ and you will now see that the stated identity is true! 
In your proof you have made a mistake in the last line. In the expression for $A$ the sum over $k$ is from $0$ to $i$, not $0$ to $n$. Now the identity follows by simply switching the order of summation. 
A: Both $\ A\ $ and $\ D: P_n(\mathbb{R}) \to P_n(\mathbb{R})\ $
are linear. Hence
$$ \frac I{0!}+\frac D{1!}+\ldots+ \frac{D^n}{n!} $$
which formally is $\ \sum_{k=0}^n \frac{D^k}{k!},\ $ is linear too.
Thus, it is enough to verify that
$$ \forall_{m=0}^n\qquad
    \left(\sum_{k=0}^n\frac{D^k}{k!}\right)\left(x^m\right)\ =
       \ (x+1)^m $$
Obviously, this is equivalent to replacing the occurence of
symbol $\ n\ $ by $\ m\ $ on the top of the $\ \sum\ $ symbol:
$$ \forall_{m=0}^n\qquad
    \left(\sum_{k=0}^m\frac{D^k}{k!}\right)\left(x^m\right)\ =
       \ (x+1)^m $$
And indeed,
$$ \forall_{m=0}^n\quad
   \left(\sum_{k=0}^m\frac{D^k}{k!}\right)\left(x^m\right)\,\ =
      \,\ \sum_{k=0}^m\, \binom mk\cdot{x^{m-k}}\,\ =
       \,\ (x+1)^m $$
That's all.
