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    $\begingroup$ $\tau(n)=O(n^{6+\epsilon})$ for all $\epsilon>0$ follows relatively simply from basic complex analysis; this would be covered in a first course in modular forms. $\endgroup$
    – user30035
    Commented Mar 29, 2013 at 9:15
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    $\begingroup$ [but the methods are very specific to modular forms; they need the relationship between $\tau(z)$ and $\tau(-1/z)$] $\endgroup$
    – user30035
    Commented Mar 29, 2013 at 9:17
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    $\begingroup$ J.-P. Serre wrote a paper on $\prod_m(1-q^m)^n$ for various other values of $n$; these are still linked to modular forms, but the behaviour of the coefficients can be rather different (sometimes almost all of the coefficients are zero!). Glasgow Math. J. 27 (1985), 203–221. You might find it of interest. $\endgroup$
    – user30035
    Commented Mar 29, 2013 at 9:24
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    $\begingroup$ The Dirichlet series $L_a(s)=\sum_{n \geq 1} \tau_a(n) n^{-s}$ can be expressed as a Mellin transform and the assumption $cr \leq a_r \leq C_r$ should imply that this Mellin transform makes sense for $\operatorname{Re}(s)$ large enough. However it's not clear to me what can be deduced on the abscissa of convergence, let alone on a polynomial bound an $\tau_a(n)$. As you explain the function is not modular in general so it's not clear what one should expect. $\endgroup$
    – François Brunault
    Commented Mar 29, 2013 at 9:42
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    $\begingroup$ Great question, and nice comments. Just a little point of history: the estimate $\tau(n)=O(n^{6+\epsilon})$ and more generally similar estimates for coefficients of cuspidal modular forms, are due to Hecke, and therefore much predate not only Deligne's but even Rankin. Also, as wccanard points out, those estimate are quite "elementary" in the sense that they use only very basic complex analysis (Cauchy's formula) plus the fact that $q \prod (1-q^n)^24$ is indeed modular, which is not trivial but was already known to Jacobi. But you mean "elementary" as "without complex analysis" here. $\endgroup$
    – Joël
    Commented Mar 29, 2013 at 14:30