Show that the function $f(x)=e^{-\frac{1}{x}},\text{ if }x>0\wedge f(x)=0,\text{ if }x\leq 0$ is smooth and all derivatives vanish in $x=0$ I had an idea but I think it is wrong. I have said $e^z$ is smooth because the n-th derivative of $e^z$ is always $e^z$ and then I have substituted $z$ with $-\frac{1}{x}$. But just calculating the first derivative Shows me that I am probably wrong because:
$(e^{z})^{'}=e^{z}=e^{-\frac{1}{x}}$
But 
$(e^{z})^{'}=(e^{-\frac{1}{x}})^{'}=\frac{e^{-\frac{1}{x}}}{x^2}$
Then also 
$e^{-\frac{1}{x}}=\frac{e^{-\frac{1}{x}}}{x^2}\iff 1=\frac{1}{x^2}$
Which is not true.
Why does the Substitution not work?
I have also tried to see a pattern in the derivatives to give an explicit Formula for the n-th derivative.
Using the product-formula for derrivatives
$$(e^{-\frac{1}{x}})^{'}=(e^{-\frac{1}{x}})(-\frac{1}{x})^{'}\overset{'}{\rightarrow}(e^{-\frac{1}{x}})^{'}(-\frac{1}{x})^{'}+(e^{-\frac{1}{x}})(-\frac{1}{x})^{''}=(e^{-\frac{1}{x}})^{'}(-\frac{1}{x})^{'}+(e^{-\frac{1}{x}})(-\frac{2}{x^3})=(e^{-\frac{1}{x}})^{'}(-\frac{1}{x})^{'}+(e^{-\frac{1}{x}})(\frac{1}{x^2})(-\frac{2}{x})=(e^{-\frac{1}{x}})^{'}(-\frac{1}{x})^{'}+(e^{-\frac{1}{x}})^{'}(-\frac{2}{x})=(e^{-\frac{1}{x}})^{'}((-\frac{1}{x})^{'}+(-\frac{2}{x}))\rightarrow ...$$
I have got the hunch that the n-th derivative can be written in a recursive form like
$$(e^{-\frac{1}{x}})^{(n)}\text{[ It is the derrivatove not the nth power ]}=(e^{-\frac{1}{x}})^{(n-1)}\cdot r$$
$r$ is the remainder and I am trying to figure out how it would look like. From my Observations so far I guess that it Looks like a sum $\sum_{k=1}^{n}(-1)^k\frac{n-k}{x^k}$. But I Need some help to prove it. 
My attempts so far have proved to be not very fruitful that's why I am asking for your support.
In order to Show that the derivative vanishes at 0 I first of all Need to find out what the derivative for positive $x$ Looks like. Then the plan would be to Show that the Limit ffrom the left as well as from the Right side is $0$

When $x<0$ then $f^{(n)}(x)=0$ for all $n\geq0$. When $x>0$ then $f^{(n)}(x)=p_n(1/x)e^{-1/x}$ for some polynomial $t\mapsto p_n(t)$. The latter is true for $n=0$ with $p_0(t)\equiv1$, and
$$f^{(n+1)}(x)={d\over dx}\biggl(p_n(1/x)e^{-1/x}\biggr)=\bigg({-1\over x^2}{p_n}'(1/x)+p_n(1/x){1\over x^2}\biggr)e^{-1/x}=:p_{n+1}(1/x)e^{-1/x}$$
for $x>0$ and $n\geq0$. Finally I claim that $f^{(n)}(0)=0$ for all $n\geq0$, and that $f^{(n)}$ is continuous at $0$. For $n=0$ this is true by definition of $f$. Assume that it is true for some $n\geq0$. Then
$$\lim_{x\to0+}{f^{(n)}(x)-f^{(n)}(0)\over x}=\lim_{x\to0+}{1\over x} p_n(1/x)e^{-1/x}=0\ ,$$
and trivially $$\lim_{x\to0-}{f^{(n)}(x)-f^{(n)}(0)\over x}=0\ .$$
This proves that $f^{(n+1)}(0)=0$.
