If $f\in\mathcal{C}^{\infty}$ then we can certainly write a Taylor series for $f$ However, analyticity requires that this Taylor series actually converge (at least across some radius of convergence) to $f$ It is not necessary that the power series for $f$ converge to $f$ as the following example shows.

Let $$ f(x)=\begin{cases} \e & x \ne 0 \\ 0 & x = 0 \end{cases}. $$ Then $f\in \mathcal{C}^{\infty}$ and for any $n\ge 0$ $f^{(n)}(0)=0$ (see below). So the Taylor series for $f$ around 0 is 0; since $f(x)>0$ for all $x\ne 0$ clearly it does not converge to $f$

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Let $p(x), q(x)\in \Reals[x]$ be polynomials, and define $$ g(x)=\frac{p(x)}{q(x)} \cdot f(x). $$ Then, for $x\ne 0$ $$ g'(x) = \frac{(p'(x) + p(x)\frac{2}{x^3})q(x) - q'(x)p(x)}{q^2(x)} \cdot\e. $$ Computing (e.g. by applying L'Hôpital's rule), we see that $g'(0)=\lim_{x\to 0}g'(x)=0$

Define $p_0(x)=q_0(x)=1$ Applying the above inductively, we see that we may write $f^{(n)}(x)=\frac{p_n(x)}{q_n(x)}f(x)$ So $f^{(n)}(0)=0$ as required.