Calculating the divisor, known to be small, of two Stirling approximations of the logarithmic Gamma function without overflows

Earlier, I asked a question on MathOverflow regarding how one might analytically approximate a function of the form: $f(n) = \prod_{i=1}^{n-1} (1-ai)$ for $a \ge 0$, $(ai) < 1$, and $n > 10^5$ or $10^6$. Robert Israel answered the question in a very nice way, observing that:

$f(n) = \dfrac{a^{n-1} \Gamma(1/a)}{\Gamma(1-n+1/a)}$

...and that one could use Stirling's series for an asymptotic approximation of $\ln(\Gamma(1-n+ 1/a))$.

My question here is: how might one actually compute values for this function (using, for example, Mathematica) for very large $n$ ($n > 10^9$ or so) and very small $a$ ($a < 10^{-30}$), without overflows/underflows, provided that the known output for $f(n)$ falls in the range of, say, $10^{-12} \le f(n) \le 1$? Are there any simple strategies for achieving this?

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Since this is a common problem many software implemented a lngamma function.
In the case of Mathematica it is named LogGamma[].

If this fails you may use an expansion of $\ln(\Gamma(x))$. You may find these at Wikipedia.

Of course once the LogGamma obtained you shouldn't compute directly the exponential ! Instead evaluate the $\log$ of the complete expression and take the exponential of the (simplified) result.

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Thanks for the suggestion... however, even the LogGamma function fails to avoid overflows/underflows. –  Roger S. Aug 16 '12 at 15:07
@RogerS.: yes of course but in a much enlarged space ! Which is the value in trouble ? –  Raymond Manzoni Aug 16 '12 at 15:10
@Roger, do you have an example? –  Guess who it is. Aug 16 '12 at 15:12
@Roger: You're using LogGamma[] wrong. Recall the identity $\log(p/q)=\log\,p-\log\,q$... –  Guess who it is. Aug 16 '12 at 15:51
@Roger, With[{n = 10^6, a = 10^-30}, Block[{\$MaxExtraPrecision = 60}, N[(n - 1) Log[a] + LogGamma[1/a] - LogGamma[1 - n + 1/a], 25]]] –  Guess who it is. Aug 16 '12 at 16:23