1. **Problem statement:** Prove that $$\log_2 \left[ \log_2 \left\{ \log_3 \left[ \log_3 27^3 \right] \right\} \right] = 0.$$\n\n2. **Recall the logarithm rules:**\n- $\log_a (b^c) = c \log_a b$\n- $\log_a a = 1$\n- $\log_a 1 = 0$\n\n3. **Evaluate the innermost logarithm:**\n$$\log_3 27^3 = \log_3 (27)^3 = 3 \log_3 27.$$\nSince $27 = 3^3$,\n$$\log_3 27 = \log_3 3^3 = 3 \log_3 3 = 3 \times 1 = 3.$$\nTherefore,\n$$\log_3 27^3 = 3 \times 3 = 9.$$\n\n4. **Next logarithm:**\n$$\log_3 [\log_3 27^3] = \log_3 9.$$\nSince $9 = 3^2$,\n$$\log_3 9 = \log_3 3^2 = 2 \log_3 3 = 2 \times 1 = 2.$$\n\n5. **Next logarithm:**\n$$\log_2 \{ \log_3 [\log_3 27^3] \} = \log_2 2 = 1.$$\n\n6. **Final logarithm:**\n$$\log_2 [ \log_2 \{ \log_3 [\log_3 27^3] \} ] = \log_2 1 = 0.$$\n\n**Answer:** The expression equals $0$, as required.