Understanding a step in a computation involving dual basis and permutations 
Since $\varphi_{i_j} \in \mathcal{T}(\mathbb{R}^n)$, for every $j = 1, \dots, k$, we have
\begin{align*}
\varphi_{i_1}\wedge\dots\wedge\varphi_{i_k}(e_{i_1}, \dots, e_{i_k}) &= \frac{k!}{1!\dots 1!}\operatorname{Alt}(\varphi_{i_1}\otimes\dots\otimes\varphi_{i_k})(e_{i_1}, \dots, e_{i_k})\\
&= \sum_{\sigma \in S_k}\operatorname{sgn}(\sigma)\varphi_{i_1}(e_{\sigma(i_1)})\dots\varphi_{i_k}(e_{\sigma(i_k)})\\
&= 1.
\end{align*}

I don't understand the final line. Why is $\phi_{i_1}(e_{\sigma(i_1)}) = 1$ for example? I thought that $\phi_{i_1}(e_{\sigma(i_1)}) = 1$ if $\sigma(i_1) = i_1$.
 A: As you point out, $\varphi_{i_1}(e_{\sigma(i_1)}) = 1$ if and only if $\sigma(i_1) = i_1$, and it is zero otherwise. Likewise, for every $j = 1, \dots, k$, $\varphi_{i_j}(e_{\sigma(i_j)}) = 1$ if and only if $\sigma(i_j) = i_j$, and it is zero otherwise. 
Now note that the product $\varphi_{i_1}(e_{\sigma(i_1)})\dots\varphi_{i_k}(e_{\sigma(i_k)})$ is zero if any of the factors are zero. The only way that the product is not zero is if all of the factors are equal to $1$, which only happens if $\sigma(i_1) = 1, \dots, \sigma(i_k) = i_k$ (i.e. $\sigma$ is the identity permutation, $\operatorname{id}$). So the sum in the penultimate line has only one non-zero summand which is the one associated to the identity permutation. As the signature of the identity permutation is $1$, we see that
$$\sum_{\sigma \in S_k}\operatorname{sgn}(\sigma)\varphi_{i_1}(e_{\sigma(i_1)})\dots\varphi_{i_k}(e_{\sigma(i_k)}) = \operatorname{sgn}(\operatorname{id})\varphi_{i_1}(e_{i_1})\dots\varphi_{i_k}(e_{i_k}) = 1.$$
