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We show for any $n \\geq 2$ that if $2^n \\leq u^n(F) \\leq u^2(F) < \\infty$ then $sl_2^n(F) \\leq \\prod_{i=2}^n (\\frac{u^i(F)}{2}+1-2^{i-1})$. As a result, if $u(F)$ is finite then $sl_2^n(F)$ is finite for any $n$, a fact which was previously proven when $\\operatorname{char}(F) \\neq 2$ by Saltman and Krashen. 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