I'd like a proof verification of the following, please.
I want to prove that $H: \Bbb R \to \Bbb R$ is of class $\mathcal C^{\infty}$, with $H$ defined by: $$H(x) = \left\{\begin{align} &h(x) = e^{\frac{-1}{x^2}}\,, \qquad x\neq 0 \\ &0\,, \qquad\qquad\qquad\,\, x = 0\end{align}\right.$$
Now, I conjecture an expression for the $n^{\text{th}}$ derivative: $$H^{(n)}(x) = \left\{\begin{align} &h^{(n)}(x)\,, \qquad x\neq 0 \\ &0\,, \qquad\qquad\,\,\, x = 0\end{align}\right.$$
The truth of this conjecture will rely on $$\lim_{t\to 0}\frac{H^{(n-1)}(t)-H^{(n-1)}(0)}{t} = 0$$ so this is what I will try to prove, by induction. The case $n = 1$ is quite simple and indeed is correct, so now I assume that $$\lim_{t\to 0}\frac{H^{(k-1)}(t)-H^{(k-1)}(0)}{t} = 0\,, \forall\, 2\leq k\lt n-1$$
We use this to calculate the limit: $$\lim_{t\to 0}\frac{H^{(n-1)}(t)-H^{(n-1)}(0)}{t}= \lim_{t\to 0}\frac{H^{(n-1)}(t)-\lim_{s\to 0}\frac{H^{(n-2)}(s)-H^{(n-2)}(0)}{s}}{t} = \lim_{t\to 0}\frac{H^{(n-1)}(t)}{t}$$
From the definition of $H$, this is actually equal to $$\lim_{t\to 0}\frac{h^{(n-1)}(t)}{t}$$
This way, the problem is reduced to showing that the above limit is zero. Another easy enough inductive proof shows that $$h^{(n)}(x) = h^{(n-1)}(x) + h'(x)$$
Proceeding again by induction: $$\lim_{t\to 0}\frac{h^{(n-1)}(t)}{t} = \lim_{t\to 0}\frac{h^{(n-2)}(t) + h'(t)}{t} = 0 + 0 = 0,$$ since the induction hypothesis was $h^{(k)}(0) = 0$ for smaller $k$.
Is this correct?