Find the number of surjective linear maps from an $n$-dimensional vector space over the field with 2 elements to itself. 
Find the number of surjective linear maps from an $n$-dimensional vector space over the field with 2 elements to itself.

Here's my attempt:
So let's say I have this vector space $V$ with a basis $\{v_1, v_2, \dots, v_n\}$, and I want the number of surjective maps from V to itself. So $T \in \mathcal{L}(V)$ can map $v_1$ to any of the other $n$ vectors, $v_2$ to any of the other $n-1$....
I'm confused as to where the "two elements" parts comes into play. I'm missing something very obvious here....
 A: It's easier to transform the question to a question about matrices. Once you choose a basis for your vector space $V$, a linear map $T$ is represented uniquely by an $n \times n$ matrix $A$ whose entries are elements of $\mathbb{F}_2$ ($0$ or $1$). The linear map will be surjective iff it is injective iff it has full rank. In terms of matrices, this means that the matrices that represent surjective matrices are matrices $A$ for which $\operatorname{rank}(A) = n$. 
Let's start with some basic observations. The number of $n \times 1$ column vectors with entries in $\mathbb{F}_2$ is $2^n$ (each entry can be either $0$ or $1$). The number of elements in a vector space of dimension $k$ over $\mathbb{F}_2$ is $2^k$ because once you choose a basis $e_1,\dots,e_k$ for $V$, each element is a unique linear combination $a_1 e_1 + \dots + a_k e_k$ where $a_i \in \mathbb{F}_2$. 
A matrix $A \in M_n(\mathbb{F}_2)$ has full rank if and only if its columns are linearly independent. This means that a matrix $A$ has full rank if and only if the first column is non-zero and the $k$-th column of the matrix doesn't belong to the span of the $1,\dots,k-1$ columns for $2 \leq k \leq n$. Having this in mind, we have $2^n - 1$ options for the first column (any non-zero vector in $\mathbb{F}_2^n$), $2^n - 2$ options for the second column (any vector that doesn't belong to the span of the first column), $2^n - 2^2$ options for the third column and so on. Hence, the number of full rank $n \times n$ matrices over $\mathbb{F}_2$ is
$$ (2^n - 1)(2^n - 2) \cdots (2^n - 2^{n-1}).$$
