Prove by induction on $n$ that if $A,B$ are $n \times n$ matrices with $B$ obtained from $A$ by swapping $i^{th}$ row and $j^{th}$ row of $A$, where $1 \le i < j \le n$, then $\det(B) = -\det(A)$.
Starting with the case $n = 2$ and claim obviously holds. Then suppose inductively that $n \ge 3$ and argument holds for all $n-1$. Then for $n$, we can express $\det(B)$ as
$$\det(B) = ...+(-1)^{i+1}a_{j1}\begin{vmatrix} a_{12} & ... & a_{1n} \\ . & & . \\ a_{i2} & ... & a_{in} \\ . & & . \\ a_{n2} & ... & a_{nn} \\ \end{vmatrix} + ... +(-1)^{j+1}a_{i1}\begin{vmatrix} a_{12} & ... & a_{1n} \\ . & & . \\ a_{j2} & ... & a_{jn} \\ . & & . \\ a_{n2} & ... & a_{nn} \\ \end{vmatrix}+...$$
Here, I have written two of the terms which either doesn't contain $a_{j2}$ row or $a_{i2}$ row. Except these two terms, other terms will include both these rows. Now, if we define $S$ to be the sum of the terms that contain both $a_{j2}$ row and $a_{i2}$ row (in other words, sum of the terms except the written ones). Then by inductive hyphothesis, when we swap $a_{j2}$ row and $a_{i2}$ row in all of these terms, their sum will be $-S$, which is sum of the terms except written ones in $\det(A)$.
After this point, in order to prove the claim, we need to show that when we multiply written terms by $(-1)$, we need to have the term $$(-1)^{i+1}a_{i1}\begin{vmatrix} a_{12} & ... & a_{1n} \\ . & & . \\ a_{j2} & ... & a_{jn} \\ . & & . \\ a_{n2} & ... & a_{nn} \\ \end{vmatrix} + (-1)^{j+1}a_{j1}\begin{vmatrix} a_{12} & ... & a_{1n} \\ . & & . \\ a_{i2} & ... & a_{in} \\ . & & . \\ a_{n2} & ... & a_{nn} \\ \end{vmatrix}$$
which should imply $$(-1)^i = (-1)^{j+1}\ and\ (-1)^j = (-1)^{i+1}$$
But when both $i$ and $j$ are even or both of them are odd, this claim is clearly false. This is the point where I am stuck. I think the problem is about $S$ and $-S$ part but I couldn't find where the things went wrong.
Any correction/help on this method is appreciated, as well as better way of induction. Thank you in advance.
Here is a sketch of a proof, I'll leave you to fill in the details.
Let $A,B,n,i,j$ be as in your question, take any $k\ne i,j$, and calculate $\det(B)$ by expanding along row $k$. Then $$\eqalign{\det(B) &=(-1)^{k+1}b_{k1}\det(B_{k1})+(-1)^{k+2}b_{k2}\det(B_{k2})+\cdots (-1)^{k+n}b_{kn}\det(B_{kn})\cr &=(-1)^{k+1}a_{k1}(-\det(A_{k1}))+(-1)^{k+2}a_{k2}(-\det(A_{k2}))+\cdots (-1)^{k+n}a_{kn}(-\det(A_{kn}))\cr &=-\det(A)\ .\cr}$$ We have used the facts