I want to prove the lemma $\sum_{i=1}^{n}(a_{i} - a_{i-1}) = a_{n} - a_{0}$ as part of another theorem.
To do this I wrote out the sum up to the nth term as follows.
$$ \sum_{i=1}^{n}(a_{i} - a_{i-1}) = (a_{1} - a_{0}) + (a_{2} - a_{1}) + (a_{3} - a_{2}) + ... (a_{n-2} - a_{n-3}) + (a_{n-1} - a_{n-2}) + (a_{n} - a_{n-1})$$
Then I showed that due to the pattern every term would be cancelled out by its negative in the next iteration, except $a_{1}$ and $-a_{0}$. My question is, since I'm assuming $a_{3}$ and $-a_{n-3}$ are cancelled in terms that happen in the continuation... am I correct to make that assumption based on how the series is defined?
Is this the sort of thing I'd have to prove with induction, or is showing that the pattern leads to a cancellation of all but $a_{n}$ and $-a_{0}$ sufficient?
Whether an induction proof is needed or just a heuristic argument is sufficient, depends on the intended reader. That said, in this particular case the induction argument is even shorter than what you wrote. Namely:
As base case, we have $\sum_{j=1}^1(a_j-a_{j -1})=a_1-a_0$. Assume as inductive hypothesis that $\sum_{j=1}^n(a_j-a_{j -1})=a_n-a_0$. Then $$ \sum_{j=1}^{n+1}(a_j-a_{j -1})=\sum_{j=1}^n(a_j-a_{j -1}) +a_{n+1}-a_n=a_n-a_0+a_{n+1}-a_n=a_{n+1}-a_0, $$ and so the formula holds by induction.