That's almost the right probability -- there should be 25 factors, so you don't need the $340/365$ at the end. The last factor is $341/365$. You're right that it's going to be quite a bit of work to compute it exactly. And what's really interesting is the numerical value -- is it less than one-half? Less than ten percent? There are lots of approximate methods for figuring out answers to problems like this, and I'll outline one here.
You can rewrite the probability you're looking for in terms of factorials as
$ P = {365! \over 340! 365^{25}} $
and this is useful because there is a very useful approximation for $n!$, called Stirling's formula:
$ n! \approx \sqrt{2 \pi n} (n/e)^n $
Now, if you were to plug in $340$ or $365$ to Stirling's formula you would get two very large numbers, which a standard calculator cannot handle; but we know $P$ should be reasonably sized. The trick here is to take logarithms. Stirling's formula can be rewritten as
$ \log n! \approx {1 \over 2} \log {2 \pi} + \left( n + {1 \over 2} \right) \log n - n $
and the logarithm of your original probability is
$ \log P = \log 365! - \log 340! - 25 \log 365. $
From Stirling's formula $\log 365! \approx 1792.331$ and $\log 340! \approx 1645.675$; this gives
$ \log P = 1792.331 - 1645.675 - 147.497 = -0.841 $
and so $P \approx e^{-0.841} = 0.432$.
(You might worry if there's an error in Stirling's approximation. It turns out that there is, and that the estimated value of $\log n!$ will always be lower than its true value by about $1/(12n)$. For example $\log 8! = 10.6046$ but Stirling gives ${1 \over 2} \log (2\pi) + 8.5 \log 8 - 8 = 10.5942$; these differ by $0.0104$ or about $1/96$. But the errors in using Stirling to approximate $\log 365!$ and $\log 340!$ are very small and in the same direction; they nearly cancel out.)