Step 1: Instead of using the floor-division formula, list every multiple of 5 up to 25 individually and count how many factors of 5 each contributes. This is a direct counting check on the formula-based answer.
Step 2: The multiples of 5 up to 25 are $5, 10, 15, 20, 25$. That is 5 numbers, so each contributes at least one factor of 5, giving a baseline of 5 factors of 5.
Step 3: Among these, $25 = 5^2$ contributes an EXTRA factor of 5 beyond its first one, since $25$ itself is a multiple of $5^2$.
Step 4: Total power of 5 dividing $25!$ $= 5$ (baseline, one per multiple of 5) $+ 1$ (extra from $25$) $= 6$.
Step 5: Since $25!$ has far more factors of 2 than factors of 5, the number of trailing zeros is limited by the power of 5, i.e. 6.
\[\boxed{6}\]