Actually, the answer in your link confused me a bit. But here is an alternative derivation (somehow similar to/same as? the answer in your link).
Look at the sequence of cubes $s_{1,n,3}=1^3+2^3+\cdots+n^3$ and the shifted sequence $s_{2,n+1,3}=2^3+\cdots+(n+1)^3$. Subtracting the former sequence from the latter obviously yields $(n+1)^3-1$. At the same time, the result of the subtraction, $s_{2,n+1,3}-s_{1,n,3}$ is given by
$$ \sum_{i=1}^n \left((i+1)^3-i^3\right) = \sum_{i=1}^n \left(3i^2+3i+1\right)=3s_{1,n,2}+3s_{1,n,1}+n.$$
In other words,
$$(n+1)^3-1=s_{2,n,3}-s_{1,n,3}=3s_{1,n,2}+3s_{1,n,1}+n.$$
Assuming that you know that $s_{1,n,1}=\sum_{i=1}^n i= \frac{n(n+1)}{2}$ you can solve for $s_{1,n,2}=\sum_{i=1}^n i^2$, which is what you have been looking for.
EDIT This can be generalized as follows (and this maybe gives a better motivation).
Suppose we wish to compute $\sum_{i=1}^n i^M$ for some positive integer $M$, and we want to find a recursive formula that expresses $\sum_{i=1}^n i^M$ in terms of $\sum_{i=1}^n i^{M-1}$, $\sum_{i=1}^n i^{M-2}$, etc.
The key idea is to look at the sum $S=\sum_{i=1}^n (i+a)^M$, for some positive integer $a$. By the binomial theorem, $(i+a)^M=\sum_{k=0}^{M}\binom{M}{k}i^ka^{M-k}$, so
$$\sum_{i=1}^n (i+a)^M=S=\sum_{k=0}^{M}\Bigl(\binom{M}{k}a^{M-k}\sum_{i=1}^n i^k\Bigr)=\binom{M}{0}a^M\sum_{i=1}^n i^0+\cdots+\binom{M}{M}a^0\sum_{i=1}^n i^{M}.$$
Conversely, notice how $S$ is just the 'right-shifted' (by $a$) analogue of $\sum_{i=1}^n i^{M}$; thus
$$\sum_{i=1}^n (i+a)^M-\sum_{i=1}^n i^M=\underbrace{(n+a)^M+(n+a-1)^M+\cdots+(a+1)^M-n^{M}-(n-1)^{M}-\cdots-1^M}_{=B}.$$
In other words,
$$\sum_{k=0}^{M-1}\Bigl(\binom{M}{k}a^{M-k}\sum_{i=1}^n i^k\Bigr)=S-\sum_{i=1}^n i^M=B$$
and this allows us to write any sum of the form $\sum_{i=1}^n i^k$ in terms of the 'remaining' sums $\sum_{i=1}^n i^q$, for $q\neq k$. In particular, we may obtain an induction formula.