Consider the "tower of power" $2^{2^{2^{.^{.^{.^2}}}}}$, where there are $2007$ twos including the base.  What is the last (units) digit of this number?

$\textbf{(A) }0\hspace{14em}\textbf{(B) }1\hspace{14em}\textbf{(C) }2$

$\textbf{(D) }3\hspace{14em}\textbf{(E) }4\hspace{14em}\textbf{(F) }5$

$\textbf{(G) }6\hspace{14em}\textbf{(H) }7\hspace{14em}\textbf{(I) }8$

$\textbf{(J) }9\hspace{14.3em}\textbf{(K) }2007$
1. To determine the last digit of the "tower of power" \(2^{2^{2^{.^{.^{.^2}}}}}\) with 2007 twos, we need to find the units digit of this large number. The units digits of powers of 2 cycle every 4 numbers: 
   \[
   \begin{align*}
   2^1 &\equiv 2 \pmod{10}, \\
   2^2 &\equiv 4 \pmod{10}, \\
   2^3 &\equiv 8 \pmod{10}, \\
   2^4 &\equiv 6 \pmod{10}, \\
   2^5 &\equiv 2 \pmod{10}, \\
   2^6 &\equiv 4 \pmod{10}, \\
   2^7 &\equiv 8 \pmod{10}, \\
   2^8 &\equiv 6 \pmod{10}, \\
   \end{align*}
   \]
   and so on.

2. Therefore, the units digit of \(2^n\) depends on \(n \mod 4\). Specifically:
   \[
   \begin{align*}
   n \equiv 0 \pmod{4} &\implies 2^n \equiv 6 \pmod{10}, \\
   n \equiv 1 \pmod{4} &\implies 2^n \equiv 2 \pmod{10}, \\
   n \equiv 2 \pmod{4} &\implies 2^n \equiv 4 \pmod{10}, \\
   n \equiv 3 \pmod{4} &\implies 2^n \equiv 8 \pmod{10}.
   \end{align*}
   \]

3. To find the units digit of \(2^{2^{2^{.^{.^{.^2}}}}}\) with 2007 twos, we need to determine the exponent \(2^{2^{2^{.^{.^{.^2}}}}} \mod 4\) where there are 2006 twos in the exponent.

4. Notice that any power of 2 greater than or equal to \(2^2\) is congruent to 0 modulo 4:
   \[
   2^2 \equiv 0 \pmod{4}, \quad 2^3 \equiv 0 \pmod{4}, \quad 2^4 \equiv 0 \pmod{4}, \quad \text{and so on.}
   \]

5. Since the exponent \(2^{2^{2^{.^{.^{.^2}}}}}\) (with 2006 twos) is clearly greater than or equal to \(2^2\), it follows that:
   \[
   2^{2^{2^{.^{.^{.^2}}}}} \equiv 0 \pmod{4}.
   \]

6. Therefore, the units digit of \(2^{2^{2^{.^{.^{.^2}}}}}\) (with 2007 twos) is determined by \(2^0 \mod 10\):
   \[
   2^0 \equiv 1 \pmod{10}.
   \]

7. However, we need to correct this step. Since \(2^0 \equiv 1 \pmod{10}\) is incorrect, we should consider the correct cycle:
   \[
   2^4 \equiv 6 \pmod{10}.
   \]

8. Thus, the correct units digit of \(2^{2^{2^{.^{.^{.^2}}}}}\) (with 2007 twos) is:
   \[
   2^{2^{2^{.^{.^{.^2}}}}} \equiv 6 \pmod{10}.
   \]

The final answer is \(\boxed{6}\).