Deﬁne the sequence of positive integers $a_n$ recursively by $a_1=7$ and $a_n=7^{a_{n-1}}$ for all $n\geq 2$.  Determine the last two digits of $a_{2007}$.
To determine the last two digits of \(a_{2007}\), we need to find \(a_{2007} \mod 100\). Given the recursive sequence \(a_1 = 7\) and \(a_n = 7^{a_{n-1}}\) for \(n \geq 2\), we will analyze the periodicity of \(7^n \mod 100\).

1. **Calculate the first few powers of 7 modulo 100:**
   \[
   \begin{aligned}
   7^1 &\equiv 7 \mod 100, \\
   7^2 &\equiv 49 \mod 100, \\
   7^3 &\equiv 343 \equiv 43 \mod 100, \\
   7^4 &\equiv 7 \times 43 = 301 \equiv 1 \mod 100.
   \end{aligned}
   \]
   We observe that \(7^4 \equiv 1 \mod 100\). This indicates that the powers of 7 modulo 100 repeat every 4 terms.

2. **Determine the periodicity:**
   Since \(7^4 \equiv 1 \mod 100\), we have:
   \[
   7^{4k} \equiv 1 \mod 100 \quad \text{for any integer } k.
   \]
   Therefore, the sequence \(7^n \mod 100\) is periodic with a period of 4.

3. **Analyze the sequence \(a_n\):**
   Given \(a_1 = 7\), we need to find the value of \(a_{2007} \mod 4\) to determine the position within the period:
   \[
   a_1 = 7 \equiv 3 \mod 4.
   \]
   Since \(a_n = 7^{a_{n-1}}\), we need to find \(7^3 \mod 4\):
   \[
   7 \equiv 3 \mod 4 \implies 7^3 \equiv 3^3 = 27 \equiv 3 \mod 4.
   \]
   Thus, \(a_2 \equiv 3 \mod 4\). By induction, we can see that \(a_n \equiv 3 \mod 4\) for all \(n \geq 1\).

4. **Determine the last two digits of \(a_{2007}\):**
   Since \(a_{2007} \equiv 3 \mod 4\), we need to find \(7^3 \mod 100\):
   \[
   7^3 \equiv 43 \mod 100.
   \]
   Therefore, the last two digits of \(a_{2007}\) are 43.

The final answer is \(\boxed{43}\).