For each positive integer $k$, we let $F(k)$ denote the number of positive integers less than or equal to $2007$ which are divisible by $k$. Clearly, $F(k) = \lfloor \frac{2007}{k} \rfloor$, the integer part of the rational number $\frac{2007}{k}$.
Because $5^5 > 2007$, the exponent of $5$ in the decimal expansion of $2007!$ is equal to
$$F(5) + F(5^2) + F(5^3) + F(5^4).$$
An easy calculation gives $F(5) = 401$, $F(25) = 80$, $F(125) = 16$ and $F(625) = 3$, hence the exponent of $5$ in the decimal expansion of $2007!$ is equal to $500$.

On the other hand, there are $1003$ even numbers among the positive integers less than $2007$ which implies that the exponent of $2$ in the decimal expansion of $2007!$ is at least $1003$. This proves that $10^{500}$ divides $2007!$, but $10^{501}$ does not divide $2007!$. The decimal expansion of $2007!$ ends in $500$ zeros.

The idea for the calculation of the digit directly in front of these $500$ zeros is the following. We calculate modulo $10$ the product of all odd factors in the product $2007!$, discarding those which are divisible by $5$. The product of all even factors is equal to $1003! \times 2^{1003}$ and we calculate modulo $10$ the product of all those factors in $1003!$ which are co-prime to $5$. From the factors not considered so far, we split off the factor $5$ and proceed as before. After repeating this procedure a few times we obtain $2007! = n \times 2^{1003} \times 5^{500}$. The calculation gives the value of $n$ modulo $10$.
The calculations modulo $10$ simplify if we take into account that for each integer $k$ we have
$$(10k + 1)(10k + 3)(10k + 7)(10k + 9) \equiv -1 \quad \mod 10 \quad (1)$$
and
$$\prod_{\substack{i=1 \\ i \neq 5}}^{9} (10k + i) \equiv -6 \quad \mod 10. \quad (2)$$
Moreover, $(-6)^m \equiv (-1)^m 6 \quad \mod 10$.

The product of all odd numbers $1, 3, ..., 1999$, co-prime to $5$, consists of $200$ products as in (1), so this contributes $(-1)^{200} \equiv 1 \mod 10$. We also have $2001 \times 2003 \times 2007 \equiv 1 \mod 10$. As seen above, there are $F(5) = 401$ positive integers below $2007$ which are divisible by $5$. Exactly $201$ of them are odd. Hence, the product of all positive odd numbers below $2007$ is equal to
$$M_1 \times 5^{201} \times 401!!,$$
where $M_1 \equiv 1 \mod 10$ and $401!!$ denotes the product of all positive odd integers less or equal to $401$.
Similarly, $401!! = M_2 \times 5^{40} \times 79!!$ with $M_2 \equiv (-1)^{40} \equiv 1 \mod 10$. In the next step we get $79!! = M_3 \times 5^8 \times 15!!$ with $M_3 \equiv (-1)^8 \equiv 1 \mod 10$. Finally,
$$15!! = 1 \times 3 \times 5 \times 7 \times 9 \times 11 \times 13 \times 15 = M_4 \times 5^2$$
with $M_4 \equiv 1 \mod 10$. Altogether we obtained so far
$$2007!! = M \times 5^{251}$$
with $M \equiv M_1 M_2 M_3 M_4 \equiv 1 \mod 10$.

To deal with the product of all even factors in $2007!$, which is equal to $1003! \times 2^{1003}$, we study $1003!$. This product contains $100$ products as in (2), which contributes $(-1)^{1006} \equiv 6 \mod 10$. In addition, we have to consider $1001 \times 1002 \times 1003 \equiv 6 \mod 10$. The number of factors in $1003!$ which are divisible by $5$ is equal to $200$ so that
$$1003! = N_1 \times 5^{200} \times 200!$$
with $N_1 \equiv 6 \pmod{10}$.
Similarly, $200! = N_2 \times 5^{40} \times 40!$ with $N_2 \equiv (-1)^{206} \equiv 6 \pmod{10}$. Next, we obtain $40! = N_3 \times 5^8 \times 8!$ with $N_3 \equiv (-1)^{46} \equiv 6 \pmod{10}$. Finally, $8! = N_4 \times 5$ with $N_4 = 2 \times 3 \times 4 \times (-4) \times (-3) \times (-2) \equiv -6 \pmod{10}$. Therefore,
$$1003! = N \times 5^{249}$$
with $N \equiv N_1 N_2 N_3 N_4 \equiv -6 \pmod{10}$. We obtain
$$2007! = M \times N \times 5^{500} \times 2^{1003} = M \times N \times 2^{503} \times 10^{500}$$
and we know $M \times N \equiv -6 \equiv 4 \pmod{10}$. It remains to find $2^{503} \equiv 8 \pmod{10}$, which is easily seen from $2^4 \equiv 6 \pmod{10}$. We conclude that
$$$$
M $\times N \times 2^{503} \equiv 4 \times 8 \equiv 2 \pmod{10}.$

Therefore, the last non-zero digit of $2007!$ is equal to $2$.