Trailing zeroes of $\dfrac{n!}{m!}$ for $n>m$ I (as a teacher) saw in a book for $8^{th}$ grade students that the number of trailing zeroes of ${n!}\times{m!}$ is the sum of the trailing zeroes of $n!$ and $m!$. There also has been noticed that the number of trailing zeroes of $\dfrac{n!}{m!}$ ($m<n$) is their subtraction. i.e.
$$(\left\lfloor \frac{n}{5}\right\rfloor+ \left\lfloor \frac{n}{5^2}\right\rfloor+
\left\lfloor \frac{n}{5^3}\right\rfloor+\cdots)-(\left\lfloor \frac{m}{5}\right\rfloor+ \left\lfloor \frac{m}{5^2}\right\rfloor+
\left\lfloor \frac{m}{5^3}\right\rfloor+\cdots).$$
But I think this is wrong because for example $\dfrac{15!}{14!}=15$ but $3-2=1$.

Can one prove that this statement is correct if $n>m-1$? If so why this restriction is necessary?

Of course it is obvious that $\dfrac{(n+1)!}{n!}=n+1$ and the number of trailing  zeroes depend on the  number of  trailing  zeroes of the number $n+1$.

Where does this strange behavior comes from? i.e. in product of factorials we sum number of trailing zeroes but in division we should care about it?

Note: I always make mistakes in simple math calculations. Am I wrong here?
 A: The formula will be wrong lots of times.  For example, $125!/122!$ has two trailing $0$’s, while the formula suggests $3$.  The problem is that there are more $5$’s in the factorization than $2$’s.
If you calculate the corresponding formula for $2$ and take the minimum of the two values, you’ll correct the formula.
The formula that is stated counts the number of $5$'s in the factorization of the quotient $\frac{n!}{m!}$.  Usually, in factorials, $5$ is scarcer than $2$.  In the quotient, however, it is possible that $2$ becomes rarer.  Therefore, if you take the formula that counts the number of $2$'s, you'll see that, for example, $\frac{15!}{14!}$ has $(7+3+1)-(7+3+1)=0$ factors of $2$, so it has no trailing $0$'s.
A: Let $\lfloor r\rfloor$ denote the floor of $r$.
For prime $p$ and positive integer $n$, let 
$V_p(n)$ denote the largest exponent $\alpha$ 
such that $p^{\alpha} | n.$ 
Note that under this definition 
$p^{(\alpha + 1)} \not | n.$
The formula for $V_p(n!)~$ is $~\lfloor \frac{n}{p^1}\rfloor ~+~
\lfloor \frac{n}{p^2}\rfloor ~+~
\lfloor \frac{n}{p^3}\rfloor ~+~
\lfloor \frac{n}{p^4}\rfloor ~+~ \cdots
$
Using this formula, the precise formula for the # of trailing zeros that 
a number will have is $\min\{V_2(n), V_5(n)\}.$
It's easy to see that in general, $V_5(n!) \leq V_2(n!).$
I recommend against any attempt to shortcut the formula. 
For example, if $n > m$, there is no way to guarantee that 
$V_5(n!) - V_5(m!) = m - n.$
In fact, one of the ways of showing that 
$\binom{n}{k}$ is an integer, 
when $~n ~\in ~\mathbb{Z^+}$ and $k \in \{0,1,\cdots, n\}$ 
is by showing that for any prime $p$, 
$V_p(n!) \geq \{V_p(k!) + V_p([n-k]!)\}.$
Addendum 
I really didn't focusing on the internals of where the OP's computation could
lead to the wrong result.
Therefore, see also Michael Burr's subsequent comment to this answer.
