How can I prove that: $55...5 = \frac{5}{9} (10^n -1)$

Question

How can I prove that:
$$55...5 = \frac{5}{9} (10^n -1)$$
while in the left side the number is $55...5$ (with $n$ occurrences of $5$).

I tried to do it with induction, but I don't know how to present the left side with $n$...

Answer 1

\begin{align} 555 \ldots 5 &=5 \times (111 \ldots 1)\\ &=5 \big(1+10+100+\cdots 10^{n-1}\big)\\ &=5\left(\frac{10^n-1}{9} \right) \end{align}

Geometric progression?

Answer 2

Or $$55\ldots5 = 5\cdot 11\ldots1 = 5\cdot {99\ldots9\over 9} = 5\cdot {100\ldots0-1\over 9}= 5\cdot {10^n-1\over 9}$$