Finding the probability that the matrix has full rank 
There are only two entries, $0$ and $1$, over $\mathbb{Z}_2$. Thus,
  only $16$ possible $2\times 2$ matrices over $Z_2$, and $6$ of them
  have full rank:
$\begin{pmatrix}0&1\\   1&0\end{pmatrix}  \quad \begin{pmatrix}1&1\\  
> 1&0\end{pmatrix}  \quad \begin{pmatrix}0&1\\   1&0\end{pmatrix}  \quad
> \begin{pmatrix}0&1\\   1&1\end{pmatrix}  \quad \begin{pmatrix}1&1\\  
> 0&1\end{pmatrix}  \quad \begin{pmatrix}1&0\\   0&1\end{pmatrix}$
Randomly generate a $n\times n$ matrix over $\mathbb {Z}_2$ (where n
  is big, say, $1000$). What's the probability that the matrix has full
  rank?

I'd want to talk something regarding to this question. 
I found the probability is equal to:
$\dfrac{(2^n-1)(2^n-2)(2^n-2^2)...(2^n-2^{n-1})}{2^n} \tag{1}$
On other hand, what's difference between $2^n$ and $2^{n^2}$?
$\dfrac{(2^n-1)(2^n-2)(2^n-2^2)...(2^n-2^{n-1})}{2^{n^2}} \tag{2}$
Can you tell whether or not I'm wrong? 
Regards!

The second formula seems correct to me.  The rows of the matrix must be linearly independent.  The first row can be any vector but the zero vector, so there are $2^n-1$ choices.  Suppose that $k$ linearly independent rows have been added.  The next row can be any vector not in the linear span of the the first $k$ rows.  Since these form a $k-$dimensional vector space over GF(2), there are $2^n-2^k$ choices for row $k+1.$
Since there are $2^{n^2}$ possible $n\times n$ matrices, the denominator should be $2^{n^2}$.
