Let $V$ be a vector space and $V^*$ its dual. For $k \in \mathbb{N}$ and $\omega \in \bigwedge^{k}(V^{*})$, $\tau \in V^*$ then $e(\tau)\omega= \tau \wedge \omega$ is the exterior multiplication. For $X \in V$, the interior multiplication is defined as: $i(X) \omega(v_1,...,v_{k-1}) = \omega(X,v_1,...,v_{k-1})$.
Now in the lecture we had a formula which wasn't proved:
$e(\tau)i(X) + i(X)e(\tau) = \tau(X)\operatorname{Id}$.
Is there an elegant, maybe even coordinate-free way to prove this ?
I will use the fact that $$ \iota_X \left( \alpha\wedge \beta\right) = (\iota_X \alpha)\wedge \beta + (-1)^k\alpha \wedge (\iota_X \beta), $$ where $\alpha$ is $a$ $k$-form and $\beta$ any differential form (see the wikipedia article for interior product). This easily follows from the definition on terms of the form $\lambda e^{i_1}\wedge\ldots\wedge e^{i_k}$, the general case on $\Lambda^k V^*$ following by linearity.
If $\tau$ is a $1$-form and $\omega$ any differential form, we have \begin{align} \iota_X(\tau\wedge \omega) &= (\iota_X\tau)\wedge \omega + (-1)^1 \tau \wedge (\iota_X\omega) \\ &= \tau(X)\omega - \tau\wedge (\iota_X\omega). \end{align} Therefore, $$ \iota_X(\tau\wedge \omega) + \tau\wedge (\iota_X\omega) = \tau(X) \omega, $$ that is, with your notations, $$ e(\tau)\circ i(X) + i(X)\circ e(\tau) = \tau(X) \mathrm{Id}. $$