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We show that: $$ \\tau(G_1 \\square G_2) \\geq \\frac{2^{(n_1-1)(n_2-1)}}{n_1n_2} (\\tau(G_1) n_1)^{\\frac{n_2+1}{2}} (\\tau(G_2)n_2)^{\\frac{n_1+1}{2}}$$ and $$\\tau(G_1 \\square G_2) \\leq \\tau(G_1)\\tau(G_2) [\\frac{2m_1}{n_1-1} + \\frac{2m_2}{n_2-1}]^{(n_1-1)(n_2-1)}.$$ We also characterize the graphs for which equality holds. 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