A basic fact about linear functions is that they are completely determined by their values on a basis of the vector space. For a multi-linear function this means (repeating this statement for each argument) that they are determined by their values where each argument independently runs through a basis of the vector space. For a function of a matrix that is linear in the rows, it means that the function is determined by the values it takes for matrices for which each row has a single entry $1$ and all other entries $0$. Concretely if such a function is written $f(v_1,\ldots,v_n)$, the arguments being rows of a matrix $A$, then by multi-linearity
$$
f(A)=\sum_{j_1,j_2,\ldots,j_n=1}^n a_{1,j_1}a_{j_2,2}\ldots a_{n,j_n}
\, f(e_{j_1},e_{j_2},\ldots,e_{j_n}),
$$
where $e_k$ is the $k$-th standard basis vector viewed as a row.
Now we must take into account that $f$ vanishes whenever two adjacent rows are equal. This implies directly that in the above summation one can drop any terms for which $j_i=j_{i+1}$ for some $i$. But also, by a standard "polarisation" argument (namely that $g(x+y,x+y)=g(x,x)+g(x,y)+g(y,x)+g(y+y)$ for bilinear $g$, so $g(x,y)=-g(y,x)$ if in addition $g$ vanishes on equal arguments), $f$ changes sign whenever we interchange two adjacent rows. So if $j_i>j_{i+1}$ for some $i$, then we have
$$
f(e_{j_1},e_{j_2},\ldots,e_{j_n})
=-f(e_{j_1},e_{j_2},\ldots,e_{j_{i+1}},e_{j_i},\ldots,e_{j_n}),
$$
and the sequence of indices $j_1,j_2,\ldots,j_{i-1},j_{i+1},j_i,j_{i+2},\ldots,j_n$ on the right, in which $j_i$ and $j_{i+1}$ have been interchanged, has one less inversion than the sequence on the left (an inversion of a sequence being a pair of positions where the term in the left position is strictly larger than the one in the right position). (You may notice I am re-doing a proof that any permutation is a composition of adjacent transpositions; one could also use that fact to show that any permutation of the arguments of $f$ affects the value by the sign of that permutation.)
Now for any sequence $(j_1,j_2,\ldots,j_n)$ other than $(1,2,\ldots,n)$, we either find that $f(e_{j_1},e_{j_2},\ldots,e_{j_n})$ is zero, or that it is determined by a similar value of $f$ but at a sequence of indices with strictly less inversions. It follows (by induction on the number of inversions) that all such terms are determined by $f(e_1,\ldots,e_n)$ alone. Finally it was given that $f(e_1,\ldots,e_n)=1$, so $f$ is completely determined.
As a bonus, this argument gives the explicit Leibniz formula for the determinant, once you check that $f(e_{\pi_1},e_{\pi_2},\ldots,e_{\pi_n})=\operatorname{sg}(\pi)$ for any permutation $\pi$ and that $f(e_{j_1},e_{j_2},\ldots,e_{j_n})=0$ for any non-permutation $(j_1,j_2,\ldots,j_n)$.