Why is the de Rham complex elliptic?

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I need to show that the exterior derivative

$$ \mathrm{d} : \Omega^r(M) \to \Omega^{r+1}(M) $$

is an elliptic differential operator. As far as I understand it, an elliptic operator is one such that the symbol $\sigma(\mathrm{d},\xi)$ is an invertible linear map for all non-zero $\xi$. In this case, we have

$$ \sigma(\mathrm{d},\xi) = \xi \,\wedge $$

(up to a factor of $i$ perhaps), which is clearly not invertible. Indeed, the dimensions of the fibres it maps between are different! I'm sure I'm missing something blindingly obvious here, but I've searched all over the web and haven't found anything that clears this up. Thanks.