Why is the completion of the ring of germs of smooth functions isomorphic to $\mathbb{R}[[T]]$? Let $C^{\infty}$ be the canonical commutative ring on the set $\{f: \mathbb{R} \rightarrow \mathbb{R} \mid  f \text{ smooth}\}$. Let $\mathfrak{m}= \{ f \mid f(0)=0 \}$ a maximal ideal.
Consider the localization at this maximal ideal $A= C^{\infty}_{\mathfrak{m}}$. A local ring with maximal ideal the image of $\mathfrak{m}$ which I will identify with $\mathfrak{m}$.

We want to complete $A$ along the principal ideal generated by the identity $(\operatorname{id})$ and see that $\hat{A} \cong \mathbb{R}[[T]]$ the ring of formal power series.

Here is, what I did. However, there are gaps to fill and I'm not fully sure if everything I did was correct.
a) It is $(\operatorname{id}) = \mathfrak{m}$:
Let $f \in \mathfrak{m}$. Then we can write $f = \operatorname{id} \cdot g$ where $g = f/{\operatorname{id}}$ on $\mathbb{R} \backslash \{0\}$ and $g(0)=f'(0)$. This is certainly continuous, but to see that this is actually a legal factorization, I must see that $g$ is even smooth. Is this clear for some reason? However, I'm not even sure if this fact is true or I need it.
Edit: I just found out that this is in fact a special case of Hadamard's Lemma. I still need to see b).
b) Certainly $\mathbb{R}[[T]]$ embedds into $\hat{A}$ via $T \mapsto \operatorname{id}$. But I don't see why this should be surjective. My intuition tells me, that maybe in $\hat{A}$ functions get identified with their formal Taylor Series around $0$, but I'm not sure if this is true. (I would somehow try to pick presentations of an element in the $(A/(\operatorname{id})^n)$ in the fashion of a)).
 A: Apologies for dredging this post up from the depths sou probably have had the answer for a while already. Anyway For the sake of completion (Ha puns!). 
Let's first denote $I:=\{f| f(0)=0\}\subset C^{\infty}(\mathbb{R})=:\mathbb{A}$ and $x:=\mbox{id}\colon\mathbb{R}\rightarrow\mathbb{R}$. 
Now in the current case, since $I$ is maximal, we don't need to bother with $A$ at all, since $\hat{A}\simeq \hat{\mathbb{A}}$ where $\hat{\mathbb{A}}$ denotes the completion of $\mathbb{A}$ at $I$. 
So $\hat{A}$ is isomorphic to the inverse limit of the $\mathbb{A}/I^k$ as $k$ goes to $\infty$. Since $\mathbb{R}[\![x]\!]$ is also an inverse limit, of the algebras $\mathbb{R}[x]/\langle x^k\rangle$, we can use the universal property of this limit to establish the isomorphism. In other words if we can show that the algebras 
$\mathbb{A}/I^k$ are isomorphic to $\mathbb{R}[x]/\langle x^k\rangle$ we are done. 
Now we can essentially establish this using the fact that $f\in I^k$ iff $D(f)(0)=0$ for all differential operators of degree at most $k$. This follows essentially from the product rule (only if) and Taylor's theorem (if). Then we see that the map 
$$\mathbb{R}[x]\hookrightarrow \mathbb{A}\twoheadrightarrow \mathbb{A}/I^k$$
has kernel exactly $\langle x^k\rangle$. So it induces a map 
$\mathbb{R}[x]/\langle x^k\rangle\rightarrow \mathbb{A}/I^k$. On the other hand we can use the intuition you had about Taylor series. So we map 
$$\mathbb{A}\rightarrow \mathbb{R}[x]/\langle x^k\rangle$$
by $f\mapsto \left[\sum_{i=0}^k\frac{1}{k!}\frac{d^kf}{dx^k}(0)x^k\right]$. Again by the fact about $I^k$ above we see that the kernel of this map is exactly $I^k$. So we obtain the map $\mathbb{A}/I^k\rightarrow \mathbb{R}[x]/I^k$ which you check is the inverse of the previous map. Also you check that 
these isomorphisms  are compatible with the inverse limits. 
Note that indeed under the resulting isomorphism functions get identified with their formal Taylor series. Also, referring to the comment by Mariano, Borel's theorem doesn't actually come up here, but there probably is a way to prove this statement in which it does. Here it is saying that the canonical map
$\mathbb{A}\rightarrow \hat{\mathbb{A}}$ (induced by the inverse system) is actually a surjection in this case (for instance, obviously, not true if you consider polynomials instead of smooth functions). 
Sorry if the answer was too little real analytic and too algebraic by the way. 
