The log Gamma function is the tool to use. Some caution is needed concerning the precision of the result, but even for large problems like this excellent precision can be attained with Excel. (This caution applies to all statistical packages that use double-precision floating point arithmetic.)
When sampling $k=200\ 000$ items uniformly out of $n=20\ 000\ 000$ items without replacement, where $d=7000$ of those items are "defective," the chance of not obtaining any defective item is the chance that the sample comes entirely from the non-defective items, of which there are $n-d$. Because all samples of size $k$ are equally probable, we only need to count the number of samples from $n-d$ items and divide by the total number of possible samples. Binomial coefficients (by definition) count samples, whence this chance is
$$\Pr(\text{no defective items in sample}) = \frac{\binom{n-d}{k}}{\binom{n}{k}}.$$
A (well-known) formula for binomial coefficients is
$$\binom{n}{k} = \frac{n!}{k!(n-k)!} = \frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}.$$
When the numerator and denominator get large--which they quickly do--we must work with their logarithms to avoid numerical overflow on the computer. Using basic laws of logs, we obtain
$$\log\binom{n}{k} = \log \Gamma(n+1) - \log \Gamma(k+1) - \log \Gamma(n-k+1).$$
Therefore
$$\eqalign{
\log \frac{\binom{n-d}{k}}{\binom{n}{k}} &= \left[\log \Gamma(n-d+1) - \log \Gamma(k+1) - \log \Gamma(n-d-k+1)\right] \\ &- \left[\log \Gamma(n+1) - \log \Gamma(k+1) - \log \Gamma(n-k+1)\right]\\
&=\log\Gamma(n-d+1) -\log\Gamma(n-d-k+1) + \log\Gamma(n-k+1)-\log\Gamma(n+1).
}$$
Excel offers a function to compute the natural logarithm of the Gamma function (which is why I rewrote the factorials in terms of $\Gamma$), GAMMALN. (We can trust this calculation because GAMMALN is easy to compute to high accuracy using its excellent asymptotic expansion and the recurrence relation $\Gamma(n+1)=n\Gamma(n)$; I strongly suspect that's how Excel does it.) Using this function, Excel finds the log to equal $-70.364\ 725\ldots$. Exponentiating this value gives $2.760\ 500\ 8\ldots\times 10^{-31}$ as the answer. Of course one will subtract the answer from $1$ at the end in order to express the complementary chance that at least one item in the sample is defective.
Because the final log of $-70. \ldots$ involved cancellation of values as large as $\log\Gamma(n+1)\approx 3\times 10^8$, some seven digits of precision were removed from the $15$ or so digits inherent in double-precision calculations. Therefore we cannot trust the answer to any more significant figures than I have written ($15-7=8$ of them). In this case the additional digits are $2.760\ 500\ \color{red}{804}\ldots$ whereas the correct answer (computed using exact arithmetic in Mathematica) is $2.760\ 500\ \color{red}{590}\ldots$, showing that this concern with loss of precision is justified.
