We have the following Theorem and its accompanying proof from pages 46-47 of Hormander's The Analysis of Linear Partial Differential Operators.
Theorem 2.3.4 If $F$ is a distribution of order $k$ with support equal to $\{0\}$, then for $\phi \in C^k$, $F$ is of the form:
$$ F(\phi) = \sum_{|\alpha| \leq k}c_{\alpha}\partial^{\alpha}\phi(0) $$
Proof: Expanding $\phi$ in a taylor series gives us:
$$\phi(x) = \sum_{|\alpha| \leq k}\frac{\partial^{\alpha}\phi(0)(x)^{\alpha}}{\alpha!} + \psi(x) $$
We have that $\partial^{\alpha}\psi(0) = 0$ when $|\alpha| \leq k$, so $F(\psi) = 0$ by theorem 2.3.3. Hence, the result follows with $c_{\alpha} = F\left(\frac{(x)^{\alpha}}{\alpha!}\right) \space \space \blacksquare$
As a note, theorem 2.3.3 mentioned in the proof is the statement that if all partials of a $C^k$ function vanish on a point in the support of $F$, then $F$ acting on that function is $0$.
I have a few questions about this proof.
1) It seems as though $\psi$ is the remainder term of the taylor expansion. Is this true?
2) Why does it follow that the partials of $\psi$ vanish at 0? Does this have to do with Taylor's theorem? It is not immediately clear.
3) Why does theorem 2.3.3 not apply to the function $\frac{x^{\alpha}}{\alpha!}$? It seems to me that all partials of this function evaluated at 0, should be 0. So by theorem 2.3.3, $F\left(\frac{x^{\alpha}}{\alpha!}\right)$ should equal 0.