I am stuck on a step in a proof, so I will write out the statement and the proof (it is not too long). I would like someone to explain the last 2 steps of the proof, if possible.
Statement: Let $U,V \subset \mathbb{R}^n$ be open, and $\varphi$ a diffeomorphism from $V$ to $U$. Then $\varphi$ is orientation-preserving $\iff \operatorname{Det}(D\varphi|_y) > 0, \ \forall y \in V.$
$\textit{Proof}:$ Let $x^1,\ldots,x^n$ be the coordinates on $U$ and $y^1,\ldots,y^n$ be the coordinates on $V$. Set $\varphi = (x^1 \circ \varphi,\ldots,x^n \circ \varphi) = (\varphi^1,\ldots,\varphi^n)$. Let $\varphi^*$ denote the pullback.
We have $$\varphi^*(dx^1 \wedge \ldots \wedge dx^n)$$ $$= \varphi^*(dx^1) \wedge \ldots \wedge \varphi^* (dx^n)$$ $$= d(\varphi^*x^1) \wedge \ldots \wedge d(\varphi^* dx^n)$$ $$= d\varphi^1 \wedge \ldots \wedge d\varphi^n $$ $$= \partial_{i_1} \varphi^1 dy^{i_1} \wedge \ldots \wedge \partial_{i_n} \varphi^n dy^{i_n}$$ $$= \operatorname{Det}(D\varphi|_y) dy^1 \wedge \ldots \wedge dy^n$$
I don´t understand the last two steps here, where does ${i_1,\ldots,i_n}$ come from, and why is this equal to the determinant of the jacobian $D\varphi$?
There is an implicit summation (called Einstein's summation convention) over the repeated indices $i_1,\ldots,i_n$. The steps decompose as follows. First, for $j\in \{1,\ldots,n\}$ one has $$ d\varphi^j = \sum_{i=1}^n \partial_i\varphi^j dy^i. $$ Hence, one has $$ d\varphi^1\wedge \cdots\wedge d\varphi^n = \left(\sum_{i=1}^n \partial_i\varphi^1 dy^i\right) \wedge \cdots \wedge \left(\sum_{i=1}^n \partial_i\varphi^n dy^i\right). $$ Now, for convenience, we write this last term as follows, $$ \left(\sum_{i=1}^n \partial_i\varphi^1 dy^i\right) \wedge \cdots \wedge \left(\sum_{i=1}^n \partial_i\varphi^n dy^i\right) = \left(\sum_{i_1=1}^n \partial_{i_1}\varphi^1 dy^{i_1}\right) \wedge \cdots \wedge \left(\sum_{i_n=1}^n \partial_{i_n}\varphi^n dy^{i_n}\right), $$ and by multilinearity, we get $$ d\varphi^1\wedge \cdots\wedge d\varphi^n = \sum_{1\leqslant i_1,\ldots, i_n\leqslant n} \partial_{i_1}\varphi^1\cdots\partial_{i_n}\varphi^n \,dy^{i_1}\wedge\cdots\wedge dy^{i_n}. $$ If two indices $i_k$ and $i_l$ are equal, then $dy^{i_k}\wedge dy^{i_l} = 0$, so that one can consider indices that are pairwise distinct only in the summation. Such a mapping $k\mapsto i_k$ is then a permutation of $\{1,\ldots,n\}$, that is, an element $\sigma \in \mathfrak{S}_n$. It follows that $$ d\varphi^1\wedge \cdots\wedge d\varphi^n = \sum_{\sigma \in \mathfrak{S}_n} \partial_{\sigma(1)}\varphi^1 \cdots \partial_{\sigma(n)}\varphi^n \, dy^{\sigma(1)}\wedge \cdots\wedge dy^{\sigma(n)}. $$ Now, it is basic multilinear algebra that $dy^{\sigma(1)}\wedge\cdots\wedge dy^{\sigma(n)} = \epsilon(\sigma)\, dy^1\wedge\cdots\wedge dy^n$, where $\epsilon\colon \mathfrak{S}_n \to \{\pm 1\}$ is the signature. Finally, one gets $$ d\varphi^1\wedge \cdots\wedge d\varphi^n = \left(\sum_{\sigma \in \mathfrak{S}_n} \epsilon(\sigma) \partial_{\sigma(1)}\varphi^1\cdots\partial_{\sigma(n)}\varphi^n\right) dy^1\wedge \cdots \wedge dy^n. $$ One concludes by recognizing the definition of the determinant of the matrix of $D\varphi$ in the coordinates $(y^1,\ldots,y^n)$ and $(x^1,\ldots,x^n)$.