Problem-solving: simple induction

I will solve a simple induction problem from the course notes, following the G. Polya approach.

Course Notes, Chapter 1, exercise 7
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Use induction to prove that, for any integers m >= 2 and n >= 1,
Sum(0 to n)_(m^t) = (m^(n + 1) - 1) / (m - 1)

1. Understanding the problem

We need to show that the equality holds for arbitrary integers m and n, provided m and n satisfy the given conditions.

2. Devising a plan

The text requires this problem be solved by induction, and seeing how this involves a summation and the need to prove an equality for arbitrary values, the context looks fit for such a method. Simple induction it will be.

3. Carrying out the plan

P(n):  For all integers m >= 2, Sum(0 to n)_(m^t) = (m^(n+1) - 1) / (m - 1)
Claim: For all integers n >= 1, P(n)

Base Case n = 1

    Let m be an arbitray integer >= 2.

    Sum(0 to 1)_(m^t) = m^0 + m^1
                      = 1 + m
                      = m + 1
    
    (m^(1+1) - 1) / (m - 1) = (m^2 - 1) / (m - 1)
                            = (m - 1)(m + 1) / (m - 1)
                            = m + 1
                              
    So P(1) is true.
    
Induction Step

    Assume P(n) is true for some n >= 1.
    
    Let m be an arbitray integer >= 2.
    
    Sum(0 to n+1)_(m^t) = Sum(0 to n)_(m^t) + m^(n+1)
                (by IH) = (m^(n+1) - 1) / (m - 1) + m^(n+1)
                        = (m^(n+1) - 1 + m^(n+1) * (m - 1)) / (m - 1)
                        = (m^(n+1) - 1 + m^(n+2) - m^(n+1)) / (m - 1)
                        = (m^(n+2) - 1) / (m - 1)
                        = (m^((n+1)+1) - 1) / (m - 1)
                        
    This implies that P(n+1) is true, which means that P(n) -> P(n+1).
    Then for all integers n >= 1, P(n) is true.

Since m was always taken as an arbitrary integer >= 2, I conclude that
for all integers n >= 1 and m >= 2, Sum(0 to n)_(m^t) = (m^(n+1) - 1) / (m - 1).

4. Looking back

I only made induction on n, and let m be an arbitrary integer >= 2 every time. This covers all valid m and n, without having to worry about both. However, instead of letting m be arbitrary, I could have made induction on m at each step of the proof. However, it seemed like overkill when a more compact alternative was so readily available.