We define a "latin square" to be an $n \times n$ grid filled with the numbers $1$ through $n$ such that every number appears in every row and every column exactly once. I need to prove that in any $n \times n$ symmetric (with respect to the main diagonal) Latin square where $n$ is odd, every number 1, 2, 3, . . . , n must appear at least once on the main diagonal.
What I've tried: I have attempted a proof by contradiction, assuming that there exists some number $i$ from between $1$ and $n$ which does not exist on the diagonal. Then I attempt to show that for the first element along the main diagonal $i$ must somewhere (call this place index $k$) in the second to $nth$ place in the first row and first column respectively. The second element of the diagonal must be in the third though $(k-1)th$ or $(k+1)th$ through $nth$ place. The $jth$ diagonal element will have $i$ in the places other than those taken by $j-1$ other $i$'s. And finally for the $nth$ diagonal element, $i$ can't be in $n-1$ places, but if $i$ isn't in the diagonal, then $i$ must be in one of those places. Therefore we have reached a contradiction.
However I did not utilize $n$ being odd or the symmetric nature of the square, so I believe I am making some mistake. Any help would be much appreciated.