All equations of the form:

1⋅9 + 2 = 11
12⋅9 + 3 = 111
123⋅9 + 4 = 1111
1234⋅9 + 5 = 11111
12345⋅9 + 6 = 111111
123456⋅9 + 7 = 1111111
1234567⋅9 + 8 = 11111111
12345678⋅9 + 9 = 111111111
123456789⋅9 + 10 = 1111111111

are true.

Restated:

For all integers n greater than or equal to 0,

 n                                 n
----            i                 ----   n+1
\    (9(n-i+1)10 )  + n + 2 = 1 + \    10
/                                 /
----                              ----
i=0                               i=0

         n
        ----          i             i
n + 1 = \    ( 10 ⋅ 10  - 9(n-i+1)10 )
        /     
        ----
        i=0

         n
        ----   i
n + 1 = \    10 (1 - 9n + 9i)
        /
        ----
        i=0

Proof:

First of all, I'm assuming that we know that:

999...total of n nines...99 + 1 = 100...total of n zeroes...0

which can be restated as

           n-1
  n        ----     i
10   = 1 + \    9⋅10 
           /
           ----
           i=0

So.

Assuming:

         n
        ----   i
n + 1 = \    10 (1 - 9n + 9i)
        /
        ----
        i=0
         n

We want to prove:

        n+1
        ----   i
n + 2 = \    10 (1 - 9(n + 1) + 9i)
        /
        ----
        i=0

We know:

                 n+1
           n+1   ----       i
1 = 10 ⋅ 10    - \    9 ⋅ 10
                 /
                 ----
                 i=0

Adding n+1 to both sides and doing some substitutions, we get:

                                 n+1
          n+1                    ----       i
n + 2 = 10   (1-9n+9n+9) + n+1 - \    9 ⋅ 10
                                 /
                                 ----
                                 i=0

                              n                    n+1
          n+1                ----   i              ----       i
n + 2 = 10   (1-9n+9(n+1)) + \    10 (1-9n+9i)  -  \    9 ⋅ 10
                             /                     /
                             ----                  ----
                             i=0                   i=0

                                                    i
Since the first term on the right is of the form '10 (1-9n+9i)' with i=n+1, this is equivelent to:

        n+1                   n+1
        ----   i              ----       i
n + 2 = \    10 (1-9n+9i)  -  \    9 ⋅ 10
        /                     /
        ----                  ----
        i=0                   i=0

        n+1
        ----   i
n + 2 = \    10 (1 - 9n + 9i - 9)
        /
        ----
        i=0

        n+1
        ----   i
n + 2 = \    10 (1 - 9n - 9 + 9i)
        /
        ----
        i=0

        n+1
        ----   i
n + 2 = \    10 (1 - 9(n + 1) + 9i)
        /
        ----
        i=0

which is what we set out to prove. QED.

-jwgh