1. **State the problem:** Simplify the expression $\log_{2025} 2025 - \log_3 \left( \frac{1}{9} \right)$.\n\n2. **Recall the logarithm properties:**\n- $\log_a a = 1$ for any base $a$.\n- $\log_a \left( \frac{1}{b} \right) = -\log_a b$.\n- $\log_a b^c = c \log_a b$.\n\n3. **Evaluate the first term:**\nSince the base and the argument are the same, $\log_{2025} 2025 = 1$.\n\n4. **Evaluate the second term:**\nRewrite $\frac{1}{9}$ as $9^{-1}$.\nSo, $\log_3 \left( \frac{1}{9} \right) = \log_3 9^{-1} = -1 \cdot \log_3 9$.\n\n5. **Calculate $\log_3 9$:**\nSince $9 = 3^2$, $\log_3 9 = \log_3 3^2 = 2$.\n\n6. **Substitute back:**\n$\log_3 \left( \frac{1}{9} \right) = -1 \times 2 = -2$.\n\n7. **Combine the terms:**\n$\log_{2025} 2025 - \log_3 \left( \frac{1}{9} \right) = 1 - (-2) = 1 + 2 = 3$.\n\n**Final answer:** $3$