Consider $GL_{n}(K)$ over field. By Dieudonne, $\forall A \in GL_{n}(K) \ (A = B*D(x))$ where $B$ is multiplication of some transvections and $D(x)=$
\begin{bmatrix} 1 & 0 &0& \dots \\ 0 & 1&0& \dots & \\ \vdots & & \dots & \\ 0 & & \dots & x \end{bmatrix}
-diagonal matrix with single last element, wich can be non unit. Its true, beacause invertable matrix has non-zero element on some $i$ position for each column $j$, we can add row $i$ to row $j+1$ and multiply by transvection $t_{j,j+1}(a_{j+1,j}^{-1}*(1-a_{j,j}))$ on the left, getting 1 on $a_{j,j}$ and subtracting multiples of $j$ row from other rows. And continue by induction untill last column, where $a_{n,n}$ can't be zero (otherwise other elements in column $n$ are zero, or we can express column $n$ as combination of other columns, each case mean, that $A$ isn't invertable), and finally we subtract multiples of $n$ row from other rows, getting result. As soon as $D:GL_{n}(K) \rightarrow K$ is bijection, we can define determinant as $x$.
What i can't understand - is how to proove that such $x$ is unique.