Yes, it's not trivial because this is about derangements. Total number of choices is $24^{25}$. For everyone to get one vote, we need a permutation, but no one is allowed to vote on himself/herself. This is a derangement, and the number of possible derangements for $n$ is shown with $!n$. So, the answer is $$\frac{!25}{24^{25}}$$
The number of derangements is calculated several ways as laid out in the wikipedia entry. One of them is the simple recursive formula
$$!n=(n-1)(!(n-1)+!(n-2))\rightarrow a_n=(n-1)(a_{n-1}+a_{n-2})$$
where $a_n$ is the number of derangements.
According to this series, we have results up to $a_{23}$, but you can easily calculate $a_{25}$.
We can test this for smaller numbers, i.e. for $n=6$, the probability is $$\frac{!6}{5^6}=\frac{265}{15625}\approx 0.017$$
An example matlab program to demostrate this is as follows:
n = 1000000;
c = 0;
m = 6; % num people
ids = 1:m;
for i = 1:n
votes = randi(m-1,1,m);
inds = votes >= ids;
votes(inds) = votes(inds) + 1;
if length(unique(votes)) == m
c = c + 1;
end
end
disp(c / n)
The output is:
0.0170