How to proof this formula

[Akmal06]

Yesterday, I faced a problem that required me to count the sum of a geometric sequence modulo $M$, $M \in \mathbb{N}$. The sequence's $ratio = 10$, and $U_1 = 1$. Then we can get the sum mod $M$ from $$((10^n -1)/9) \: mod \: M$$. But, for large $n$, the formula isn't efficient. So I do searching. I got a better formula: $$(10^n \: mod \: \: (9 \times M) \: -1)/9$$
In the end, I got accepted by both formulas. But I can't prove why $$((10^n -1)/9) \: mod \: M \: = \: (10^n \: mod \: \: (9 \times M) \: -1)/9$$
Please help me to prove it. Thank you.

*Sorry for my bad English, since English is not my first language.

 

[Cubist]

Please double check these steps, since I'm not completely sure

$$\left( \frac{10^n-1}{9} \right) \bmod M$$
$$= \left( \left( 10^n-1 \right) \bmod (9M) \right) / 9$$
$$= \left( \left( \left(10^n \bmod (9M)\right) -1 \right) \bmod (9M) \right) / 9$$
Because $10^n \bmod (9M)$ can never be 0, the outer "mod" can safely be removed

$$= \left( \left(10^n \bmod (9M)\right) -1 \right) / 9$$