Let $(V,h)$ be real Euklidean vector space and $(V \otimes_{\mathbb{R}}\mathbb{C}, \overline{h})$ complex extension by complex linearity. We consider the exterior algebra space $E= \wedge^{\bullet} V= \bigoplus_{i=0} ^{\infty} \wedge^{i}V$ and it's complexification $E \otimes_{\mathbb{R}}\mathbb{C}$.
$h$ can extended in canonical way to every exterior $\wedge^k V$ by
$$h(v_1 \wedge... \wedge v_k,w_1 \wedge... \wedge w_k) := \det(h(v_i,w_j)) $$
At wikipedia page dealing with Exterior algebra I found the claim that for $, x \in V, v \in \wedge^{k-1} V, w \in \wedge^{k} V$ the operators $x \wedge -$ and $i_{x^*}= h(x,-)$ are adjoint:
$$h(x \wedge v, w)= h(v, i_{x^*} w)$$
Why that's true?
Let now $y \in V \otimes_{\mathbb{R}}\mathbb{C}$ and $\overline{y}$ the complex conjugated. Is it true & why if it is that for $v \in \wedge^{k-1} V \otimes_{\mathbb{R}}\mathbb{C}, w \in \wedge^{k} \otimes_{\mathbb{R}}\mathbb{C}$ the "conjugated" adjunction
$$\overline{h}(y \wedge v, w)= \overline{h}(v, i_{\overline{y}^*} w)$$
is still true?