Consider the function $$f(z)=e^{-1/z}$$ This function has the following Laurent series about the point zero $$\sum_{n=0}^\infty \frac{z^{-n}}{n!} $$ Laurent series by definition is $$ f(z)=\sum_{n=-\infty}^\infty a_n(z-c)^n $$ $$ a_n=\frac{1}{2\pi i} \oint_\gamma \frac{f(z)\,\mathrm{d}z}{(z-c)^{n+1}}.\, $$
For $n=0$, $$ a=\frac{1}{2\pi i} \oint_\gamma \frac{f(z)\,\mathrm{d}z}{(z-c)^{}}=f(c).\,$$
Therefore, $f(0)=1$
Is this right? Where am I wrong?
You probably remember the relation "$a_0=f(0)$" from Taylor series centered at $0$. But it does not apply to Laurent series.
Laurent series work differently: in general, they converge in an annulus, not in a disk, and therefore do not represent the function at the center point. Formally plugging in $z=0$ into the series does not even make sense when it has negative powers of $z$ in it.