Prove that the function $f(x)$ given by $$f(x) = \left\{ \begin{align} e^{-1/x} \ &\text{if} \ x > 0,\\ 0 \ &\text{otherwise}. \end{align} \right.$$ is of class $C^\infty$ as follows: Given any integer $n \ge 0$, define $f_n \colon \mathbf R \to \mathbf R$ by the equation $$f_n(x) = \left\{ \begin{align} \dfrac{e^{-1/x}}{x^n} \ &\text{for} \ x > 0,\\ 0 \ &\text{for} \ x \le 0. \end{align} \right.$$
(a) Show that $f_n$ is continuous at $0$. [Hint: Show that $a < e^a$ for all $a$. Then set $a = \dfrac{t}{2n}$ to conclude that $$\frac{t^n}{e^t} < \frac{(2n)^n}{e^{t/2}}$$ Set $t = \dfrac{1}{x}$ and let $x$ approach $0$ through positive values.]
(b) Show that $f_n$ is differentiable at $0$.
(c) Show that $f'_n(x) = f_{n + 2}(x) - nf_{n + 1}(x)$ for all $x$.
(d) Show that $f_n$ is of class $C^\infty$.
I may be missing something obvious, but I've completed the four steps and I'm not sure how the result follows.
$C^\infty$ is closed under multiplication and $f(x)=x^nf_n(x)$.