In the proof of the Liouville's Formula, there is a step that says
$$\sum_{k=1}^n \left( \sum_{\sigma\in \mathcal{S}_n} \epsilon(\sigma)\left(\sum_{l=1}^n a_l^k(t)\varphi_{\sigma(k)}^l(t) \right)\prod_{\substack{i=1 \\ i\neq k}}^n \varphi_{\sigma(i)}^{i} (t) \right)=\sum_{k=1}^n \left( \sum_{\sigma\in \mathcal{S}_n} \epsilon(\sigma)a_k^k(t)\varphi_{\sigma(k)}^k (t)\prod_{\substack{i=1 \\ i\neq k}}^n \varphi_{\sigma(i)}^{i} (t) \right)$$
I know that this is a property of determinant, but I can't see how removing terms of the sum the equality holds.
Thanks!
I assume that your $\mathcal{S}_n$ stands for the set of permutations of $\left\{1,2,\ldots,n\right\}$; let me rename this set as $S_n$.
I also assume that your $\epsilon\left(\sigma\right)$ stands for the sign of a permutation $\sigma$; let me rename it as $\left(-1\right)^\sigma$.
I shall abbreviate $a^k_l \left(t\right)$ as $c_{k, l}$, and I shall abbreviate $\varphi^k_l \left(t\right)$ as $b_{k, l}$. I assume that all the $n^2$ elements $c_{k,l}$ (for $k, l \in \left\{1,2,\ldots,n\right\}$) and all the $n^2$ elements $b_{k,l}$ (for $k, l \in \left\{1,2,\ldots,n\right\}$) belong to a commutative ring.
Thus, your equality rewrites as \begin{equation} \sum_{k=1}^n \sum_{\sigma \in S_n} \left(-1\right)^\sigma \left(\sum_{l=1}^n c_{k, l} b_{l, \sigma\left(k\right)} \right) \prod_{i \neq k} b_{i, \sigma\left(i\right)} = \sum_{k=1}^n \left( \sum_{\sigma\in S_n} \left(-1\right)^\sigma c_{k, k} b_{k, \sigma\left(k\right)} \prod_{i \neq k} b_{i, \sigma\left(i\right)}\right) , \label{darij.eq.claim} \tag{1} \end{equation} where the "$\prod\limits_{i \neq k}$" symbol stands for $\prod\limits_{\substack{i \in \left\{1,2,\ldots,n\right\};\\ i \neq k}}$.
Let me prove this equality now. Let $B$ be the $n \times n$-matrix whose $\left(i, j\right)$-th entry is $b_{i, j}$ for all $i, j \in \left\{1,2,\ldots,n\right\}$. Fix $k \in \left\{1,2,\ldots,n\right\}$.
Furthermore, let $l \in \left\{1,2,\ldots,n\right\}$ be such that $l \neq k$. Let $\overline{B}$ be the $n \times n$-matrix obtained from $B$ by replacing its $k$-th row by its $l$-th row. Thus, the $k$-th row of $\overline{B}$ equals the $l$-th row of $B$, whereas each of the other $n-1$ rows of $\overline{B}$ equals the corresponding row of $B$. Hence, the $k$-th row of $\overline{B}$ equals the $l$-th row of $\overline{B}$ (since both of these rows equal the $l$-th row of $B$). Thus, the matrix $\overline{B}$ has two equal rows (since $l \neq k$). Consequently, $\det \overline{B} = 0$. On the other hand, the definition of $\overline{B}$ shows that the $\left(i, j\right)$-th entry of the matrix $\overline{B}$ is $\begin{cases} b_{i, j}, & \text{if } i \neq k; \\ b_{l, j}, & \text{if } i = k \end{cases}$ for all $i, j \in \left\{1,2,\ldots,n\right\}$. Hence, if we compute $\det \overline{B}$ using the Leibniz formula for determinants, then we obtain \begin{equation} \det \overline{B} = \sum_{\sigma \in S_n} \left(-1\right)^\sigma \prod_{i=1}^n \begin{cases} b_{i, \sigma\left(i\right)}, & \text{if } i \neq k; \\ b_{l, \sigma\left(i\right)}, & \text{if } i = k \end{cases} = \sum_{\sigma \in S_n} \left(-1\right)^\sigma b_{l, \sigma\left(k\right)} \prod_{i\neq k} b_{i, \sigma\left(i\right)} \end{equation} (here, we have split off the factor for $i = k$ from the product). Comparing this with $\det \overline{B} = 0$, we obtain \begin{equation} \sum_{\sigma \in S_n} \left(-1\right)^\sigma b_{l, \sigma\left(k\right)} \prod_{i\neq k} b_{i, \sigma\left(i\right)} = 0 . \label{darij.eq.3} \tag{2} \end{equation}
Now, forget that we fixed $l$. Thus, we have proven the equality \eqref{darij.eq.3} for each $l \in \left\{1,2,\ldots,n\right\}$ satisfying $l \neq k$. Now, \begin{align} & \sum_{\sigma \in S_n} \left(-1\right)^\sigma \left(\sum_{l=1}^n c_{k, l} b_{l, \sigma\left(k\right)} \right) \prod_{i \neq k} b_{i, \sigma\left(i\right)} \\ &= \sum_{l=1}^n c_{k, l} \sum_{\sigma \in S_n} \left(-1\right)^\sigma b_{l, \sigma\left(k\right)} \prod_{i \neq k} b_{i, \sigma\left(i\right)} \\ &= c_{k, k} \sum_{\sigma \in S_n} \left(-1\right)^\sigma b_{k, \sigma\left(k\right)} \prod_{i \neq k} b_{i, \sigma\left(i\right)} + \sum_{l \neq k} c_{k, l} \underbrace{\sum_{\sigma \in S_n} \left(-1\right)^\sigma b_{l, \sigma\left(k\right)} \prod_{i \neq k} b_{i, \sigma\left(i\right)}}_{\substack{=0 \\ \left(\text{by \eqref{darij.eq.3}}\right)}} \\ & \qquad \left(\text{here, we have split off the addend for $l = k$ from the sum}\right) \\ &= c_{k, k} \sum_{\sigma \in S_n} \left(-1\right)^\sigma b_{k, \sigma\left(k\right)} \prod_{i \neq k} b_{i, \sigma\left(i\right)} + \underbrace{\sum_{l \neq k} c_{k, l} 0}_{= 0} \\ &= c_{k, k} \sum_{\sigma \in S_n} \left(-1\right)^\sigma b_{k, \sigma\left(k\right)} \prod_{i \neq k} b_{i, \sigma\left(i\right)} . \end{align}
Now, forget that we fixed $k$. Adding the equality we have just obtained for all $k \in \left\{1,2,\ldots,n\right\}$, we find \begin{align} \sum_{k=1}^n \sum_{\sigma \in S_n} \left(-1\right)^\sigma \left(\sum_{l=1}^n c_{k, l} b_{l, \sigma\left(k\right)} \right) \prod_{i \neq k} b_{i, \sigma\left(i\right)} &= \sum_{k=1}^n c_{k, k} \sum_{\sigma \in S_n} \left(-1\right)^\sigma b_{k, \sigma\left(k\right)} \prod_{i \neq k} b_{i, \sigma\left(i\right)} \\ &= \sum_{k=1}^n \left( \sum_{\sigma\in S_n} \left(-1\right)^\sigma c_{k, k} b_{k, \sigma\left(k\right)} \prod_{i \neq k} b_{i, \sigma\left(i\right)}\right) . \end{align} This proves \eqref{darij.eq.claim}.