Riesz representation theorem in $C^k([0,1])$ I'm trying to figure out the next exercise

Let be $k\geq 1$ and $C^k([0,1])$ the banach space of all k-times differentiable function with the norm
  $$\lVert f\rVert_{k,\infty}=\sum_{i=0}^{k}\lVert f^{(i)}\rVert_{\infty}$$
  Show that $L\in (C^{k}([0,1]),\lVert\cdot\rVert_{k,\infty})^*$ iff exists $(\mu_0,...,\mu_k)$ radon measures in $[0,1]$ such that for all $f \in C^k([0,1])$ 
  $$L(f)=\sum_{i=0}^k\int f^{(i)}d\mu_i$$

Where  $(C^{k}([0,1]),\lVert\cdot\rVert_{k,\infty})^*$ is the dual space.
There is a similar exercise in the Folland Chapter 7 exercise 27. But in this exercise I use de Taylor's formula, now I think that I can't.
Any idea?
 A: First, observe that $C^{k}([0,1])$ is isomorphic to a closed linear subspace of $C([0,1])^{k + 1}$, the isomorphism being 
$$f \mapsto (f,f',f^{(2)},\dots,f^{(k)}).$$
Henceforth, we will identify $C^{k}([0,1])$ with the image of this map.
(If you think about it, the norm $\|f\|_{C^{k}([0,1])} = \sum_{i = 0}^{k} \|f^{(i)}\|_{C([0,1])}$ is invoking this embedding implicitly.)    
If $L$ is a bounded linear functional on $C^{k}([0,1])$, then by the Hahn-Banach theorem there is a bounded linear functional $\tilde{L}$ on $C([0,1])^{k + 1}$ such that 
$$\forall f \in C^{k}([0,1]) \quad \tilde{L}(f) = L(f).$$
Since $\tilde{L}$ is a bounded linear functional on $C([0,1])^{k}$, there are bounded linear functionals $\tilde{L}_{0},\tilde{L}_{1},\dots,\tilde{L}_{k}$, each defined on $C([0,1])$, such that
$$\forall (f_{0},\dots,f_{k}) \in C([0,1])^{k} \quad \tilde{L}(f_{0},\dots,f_{k}) = \tilde{L}_{0}(f_{0}) + \dots + \tilde{L}_{k}(f_{k}).$$
By the Riesz Representation Theorem, for each $i$, we can identify $\tilde{L}_{i}$ with the action of a unique Radon measure $\mu_{i}$ on $[0,1]$.  This implies
$$\forall f \in C([0,1])^{k} \quad L(f) = \sum_{i = 0}^{k} \int_{0}^{1} f^{(i)}(x) \, \mu_{i}(dx).$$  
