There is one way to more or less predict the coefficients in one of these sums. It is not a use of induction as such, but I think it is worth knowing when you are beginning the study of induction. It is just Pascal's triangle. $ \sum_{k=0}^n \left( \begin{array}{c} k \\ 0 \end{array} \right) = \left( \begin{array}{c} n+1 \\ 1 \end{array} \right), $ $ \sum_{k=1}^n \left( \begin{array}{c} k \\ 1 \end{array} \right) = \left( \begin{array}{c} n+1 \\ 2 \end{array} \right), $ $ \sum_{k=2}^n \left( \begin{array}{c} k \\ 2 \end{array} \right) = \left( \begin{array}{c} n+1 \\ 3 \end{array} \right), $ $ \sum_{k=3}^n \left( \begin{array}{c} k \\ 3 \end{array} \right) = \left( \begin{array}{c} n+1 \\ 4 \end{array} \right), $ $ \sum_{k=4}^n \left( \begin{array}{c} k \\ 4 \end{array} \right) = \left( \begin{array}{c} n+1 \\ 5 \end{array} \right). $
So then you need expressions such as $ k = \left( \begin{array}{c} k \\ 1 \end{array} \right), $ $ k^2 = 2 \; \left( \begin{array}{c} k \\ 2 \end{array} \right) + \left( \begin{array}{c} k \\ 1 \end{array} \right), $ $ k^3 = 6 \; \left( \begin{array}{c} k \\ 3 \end{array} \right) + 6 \; \left( \begin{array}{c} k \\ 2 \end{array} \right) + \left( \begin{array}{c} k \\ 1 \end{array} \right), $ $ k^4 = 24 \; \left( \begin{array}{c} k \\ 4 \end{array} \right) + 36 \; \left( \begin{array}{c} k \\ 3 \end{array} \right) + 14 \; \left( \begin{array}{c} k \\ 2 \end{array} \right) + \left( \begin{array}{c} k \\ 1 \end{array} \right), $ $ k^5 = 120 \; \left( \begin{array}{c} k \\ 5 \end{array} \right) + 240 \; \left( \begin{array}{c} k \\ 4 \end{array} \right) + 150 \; \left( \begin{array}{c} k \\ 3 \end{array} \right) + 30 \; \left( \begin{array}{c} k \\ 2 \end{array} \right) + \left( \begin{array}{c} k \\ 1 \end{array} \right). $ It helps that these expressions still work if you take $ \left( \begin{array}{c} k \\ t \end{array} \right)= \; 0 $ whenever $ 0 \leq k < t.$ Indeed, under the same rule, the sums in the first section can all start at $k=0$ instead of the beginning values I typed in.
Well, just for laughs, I worked it through, $ \sum_{k=0}^n k^4 = 24 \; \left( \begin{array}{c} n+1 \\ 5 \end{array} \right) + 36 \; \left( \begin{array}{c} n+1 \\ 4 \end{array} \right) + 14 \; \left( \begin{array}{c} n+1 \\ 3 \end{array} \right) + \left( \begin{array}{c} n+1 \\ 2 \end{array} \right). $ Writing things out, I got a common denominator of 30, then I put everything together into one fraction and got $ \sum_{k=0}^n k^4 = \frac{6 n^5 + 15 n^4 + 10 n^3 -n}{30}, $ where the lack of a constant term was predictable but the lack of an $n^2$ term seems an accident.
Now that I have been through this one, line by line, I can see how absolutely crucial it is that I be able to correctly rewrite one of the formulas in the first section as starting at $k=0,$ as
$ \sum_{k=0}^n \left( \begin{array}{c} k \\ 4 \end{array} \right) = \left( \begin{array}{c} n+1 \\ 5 \end{array} \right), $ which is true because we take $ \left( \begin{array}{c} k \\ t \end{array} \right)= \; 0 $ whenever $ 0 \leq k < t.$