By the nature of the Laplace expansion, a proof is necessarily going to be ugly and technical.
HINT: it will be much easier to prove this yourself than to follow this proof.
Let me use the same notations as in wikipedia. So take a matrix B. Let me show that expansion along the first row yields the same result as expansion along the second row. The more general statement is left to you. We prove this by induction. For the 1x1 case, the statement is trivial.
So assume that B is nxn. Expansion along the first row yields
b_{1,1}C_{1,1}+...+b_{1,n}C_{1,n}=b_{1,1}M_{1,1}-b_{1,2}M_{1,2}+...+(-1)^{n+1}b_{1,n}M_{1,n}
Expansion along the second row yields
b_{2,1}C_{2,1}+...+b_{2,n}C_{2,n}=-b_{2,1}M_{2,1}+b_{2,2}M_{2,2}+...+(-1)^{n+2}b_{2,n}M_{2,n}
We wish to calculate M_{1,1}. By definition this is the determinant of the matrix that results if we remove the first row and the first column from B. By induction hypothesis, we can calculate this determinant by taking the Laplace expansion along the first row. So we can write
M_{1,1}=b_{2,2}D_{1,2}^{1,2} - b_{3,2}D_{1,2}^{1,3}+...+(-1)^{2+n}D_{1,2}^{1,n}
where D_{a,b}^{c,d} is the determinant of the matrix resulting from B if we remove row a and b, and if we remove column c and d.
In general:
M_{1,k}=(-1)^{\delta(1,k)} b_{2,1}D_{1,2}^{1,k} +(-1)^{\delta(2,k)} b_{2,2}D_{1,2}^{2,k}+... + (-1)^{\delta(n,k)}b_{2,n}D_{1,2}^{n,k}
We used the following notations: D_{1,2}^{k,k}=0 and \delta(l,k) is the number of elements in \{1,...,l-1\}\setminus\{k\}.
To calculate M_{2,k}, we calculate this matrix by taking the Laplace expansion along the first row. We get
M_{2,k}=(-1)^{\delta(1,k)} b_{1,1}D_{1,2}^{1,k} + (-1)^{\delta(1,k)}b_{1,2}D_{1,2}^{2,k}+...+(-1)^{\delta(n,k)}b_{1,n}D_{1,2}^{n,k}
We substitute these values of M_{1,k} and M_{2,k} into the original sum.
By definition we know that D_{1,2}^{j,k}=D_{1,2}^{k,j}. We wish to prove that the coefficients of these terms are equal.
The coefficient of D_{1,2}^{j,k} in the first sum is:
(-1)^{k+1}b_{1,k}(-1)^{\delta(j,k)}b_{2,j}
The coefficient of D_{1,2}^{k,j} in the first sum is:
(-1)^{j+1}b_{1,j}(-1)^{\delta(k,j)}b_{2,k}
So together, we have
(-1)^{k+\delta(j,k)+1}b_{1,k}b_{2,j}+ (-1)^{j+\delta(k,j)+1}b_{2,k}b_{2,j}
We do the same for the terms in the second sum. The coefficient of D_{1,2}^{j,k} in the second sum is:
(-1)^{k+2}b_{2,k}(-1)^{\delta(j,k)}b_{1,j}
The coefficient of D_{1,2}^{k,j} in the second sum is:
(-1)^{j+2}b_{2,j}(-1)^{\delta(k,j)}b_{1,k}
So together we have
(-1)^{k+\delta(j,k)+2}b_{2,j}b_{1,j}b_{2,k} + (-1)^{j+\delta(k,j)+2}b_{2,j}b_{1,k}
In order that both sums are equal, it suffices to show that
(-1)^{j+\delta(k,j)+2}=(-1)^{k+\delta(j,k)+1}
Assume first that k<j. Then \delta(k,j) is the number of elements in \{1,...,k-1\}\setminus \{j\} and this is k-1. So the left-hand side becomes
(-1)^{j+k+1}
If k<j, then \delta(j,k) is the number of elements in \{1,...,j-1\}\setminus \{k\} and this is j-2. So the right hand side becomes
(-1)^{k+j-1}
Clearly, the left-hand side equals the right-hand side.