Prove that if $A$ has linearly dependent rows or columns then $\det(A)=0$.
$$\DeclareMathOperator{\sgn}{sgn} \det(A)=\sum_{\sigma\in s_n}\Bigl(\sgn(\sigma)\prod_{i=1}^{n}a_{i,\sigma(i)}\Bigr)=\sum_{\sigma\in s_n\text{ even}}\Bigl(\sgn(\sigma)\prod_{i=1}^{n}a_{i,\sigma(i)}\Bigr)-\sum_{\sigma\in s_n\text{ odd}}\Bigl(\sgn(\sigma)\prod_{i=1}^{n}a_{i,\sigma(i)}\Bigr)$$
How to continue from here using Leibniz formula?
By linearity, your statement is equivalent to prove that determinant is zero whenever two columns/rows are the same. I'll work with columns (transpose for rows). So by definition : $$ \det(A) = \sum_{\sigma\in S_n} \mathrm{sgn}(\sigma)\prod_{i=1}^n A_{\sigma(i)i}$$ Suppose that there are some $i\neq j$ such that the $i$-th and the $j$-th columns are the same.
Denote $\tau$ the transposition of the pair $(i,j)$ and let $\sigma$ be a permutation of numbers $1,\ldots, n$.
Then $\sigma\mapsto \sigma\circ \tau$ is a bijection between $A_n$ and $S_n\setminus A_n$, where $A_n$ denotes the alterning group of even permutations, but $S_n$ is the disjoint union of $A_n$ and $S_n\setminus A_n$ so $$ \det(A) = \sum_{\sigma\in A_n} \mathrm{sgn}(\sigma)\prod_{i=1}^n A_{\sigma(i)i} + \sum_{\sigma\in S_n\setminus A_n} \mathrm{sgn}(\sigma)\prod_{i=1}^n A_{\sigma(i)i}$$ Now, using the bijection between $A_n$ and $S_n\setminus A_n$ we get $$ \det(A) = \sum_{\sigma\in A_n} \mathrm{sgn}(\sigma)\prod_{i=1}^n A_{\sigma(i)i} + \sum_{\sigma\in A_n} \mathrm{sgn}(\sigma\circ \tau)\prod_{i=1}^n A_{\sigma(\tau(i))i}$$ However $\mathrm{sgn}(\sigma\circ \tau) = \mathrm{sgn}(\sigma)\cdot \mathrm{sgn}(\tau) = -\mathrm{sgn}(\sigma) $ and for a given permutation $\sigma$ $$\prod_{i=1}^n A_{\sigma(\tau(i))i} =\prod_{i=1}^n A_{\sigma(i)i}$$ Finally you obtain the following... $$ \det(A) = \sum_{\sigma\in A_n} \mathrm{sgn}(\sigma)\prod_{i=1}^n A_{\sigma(i)i} -\sum_{\sigma\in A_n} \mathrm{sgn}(\sigma)\prod_{i=1}^n A_{\sigma(i)i} = 0$$