<s>[INST] Number of $0$ in the end of $11^n-1$ $n$ is integer, calculate number of $0$ in the end of $11^n-1$(i.e. largest integer $m$ such that $10^m|11^n-1$).
The original question was $n=100$ and I could only choose $m$ from 1 to 5. I calculated the $11^{100}-1$ mod $100000$ and got the answer, I'm looking for much better and general approach.
 [/INST] The binomial expansion used for $n=100$ in the answer to the linked question can also be applied to general $n$. Let us set
$$11^n-1= (10+1)^n-1= \sum_{i=0}^n {n \choose i} 10^i -1$$
The sum can be expanded as
$$ 11^n-1= 10 \cdot n+ 100 \cdot \frac{n(n-1)}{2!}+ 1000 \cdot \frac{n(n-1)(n-2)}{3!}....$$
Now, as correctly hypothesized by Martingalo in his comment, if $n$ is not a multiple of $10$ it is clear that the sum is divisible only by $10$ and not by higher powers of $10$. So, the quantity $11^n-1$ ends with only one zero.
On the other hand, if $n$ is a multiple of $10$, we can write it in the form $n=j \cdot 10^k$ (where $j$ is not divisible by $10$ and then $k$ is the number of times that the factor $10$ is contained in $n$). Then, in the sum above:


*

*the first term $10 \cdot n$ contains the factor $10$ exactly $k+1$ times;

*the second term $100 \cdot \frac{n(n-1)}{2!}$ also contains the factor $10$ exactly $k+1$ times (in fact, multiplying $n$ by the odd number $n-1$ and dividing by $2$ we get that the binomial coefficient ${n \choose 2}$ contains the factor $10$ one time less as compared to $n$, i.e. $k-1$ times); 

*the third term $1000 \cdot \frac{n(n-1)(n-2)}{3!}$ contains the factor $10$ exactly $k+3$ times (in fact, looking again at the binomial coefficient, we can note that multiplying $n$ by the even number $(n-1)(n-2)$ and dividing by $2 \cdot 3$,  we clearly obtain a number that contains the factor $10$ the same number of times of $n$, i.e. $k$ times;

*all successive terms contain the factor $10 $ at least $4$ times.
Therefore, taking only the first and second term of the sum, we have 
$$10 \cdot n+ 100 \cdot \frac{n(n-1)}{2!}=10n+50n^2-50n=50n^2-40n=10n(5n-4)$$
and reminding that $n$ is of the form $n=j \cdot 10^k$ with $j$ not divisible by $10$, we get
$$10n(5n-4)=10^{k+1} j(5j\cdot 10^k-1)$$
which clearly contains the factor $10$ exactly $k+1$ times. So, in this case, the whole expansion of $11^n-1$ also contains the factor $10$ exactly $k+1$ times, and then ends with $k+1$  zeros.
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