Proof by induction: showing that two sums are equal usually the tasks look like
$$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$
or
$$\sum_{i=0}^n i^2 = i_1^2 + i_2^2 + i_3^2+...+i_n^2$$
But for the following task I have this form:
$$\left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^nk^3 $$
First I am a little confused by how I approach this. Do I transform them into a complete term like in the first example? Or can I do it by just using the sums themselves? And how should I treat the square of the sum best?
The first step is pretty straight forward creating the base case. But as soon as I get into doing the "Induction Step" I am getting stuck and not sure how to proceed.
I am looking to know best practice for this case.
Edit:
This question is a little different, since it is expected to proove this only by using complete induction using the sum notation.
 A: For the step you get:
$$(\sum_{k=1}^{n+1} k)^2 = $$
$$[(\sum_{k=1}^n k) + (n+1)]^2 = $$
$$(\sum_{k=1}^n k)^2 + 2(n+1)\sum_{k=1}^n k + (n+1)^2 = $$ ($\sum_{k=1}^n k = \frac{n(n+1)}{2}$)
$$(\sum_{k=1}^n k)^2 + 2(n+1)\frac{n(n+1)}{2} + (n+1)^2 = $$
(inductive hypothesis)
$$\sum_{k=1}^n k^3 + n(n+1)^2 + (n+1)^2 =$$
$$\sum_{k=1}^n k^3 + (n+1)(n+1)^2=$$
$$\sum_{k=1}^n k^3 + (n+1)^3=$$
$$\sum_{k=1}^{n+1} k^3 $$
A: Assume that $\displaystyle\left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^nk^3$ holds for $n.$ We want to show that  $\displaystyle\left(\sum_{k=1}^{n+1} k\right)^2 = \sum_{k=1}^{n+1}k^3.$ How to do it? Note that
$$\begin{align}\left(\sum_{k=1}^{n+1} k\right)^2&=\left(\sum_{k=1}^{n} k+n+1\right)^2\\&= \color{blue}{\left(\sum_{k=1}^{n} k\right)^2}+2(n+1)\sum_{k=1}^nk+(n+1)^2\\&\underbrace{=}_{\rm{induction}\:\rm{hypothesis}}\color{blue}{\sum_{k=1}^nk^3}+\color{red}{2(n+1)\sum_{k=1}^nk+(n+1)^2}\\&=\sum_{k=1}^{n+1}k^3\end{align}$$ if and only if $\displaystyle(n+1)^3=2(n+1)\sum_{k=1}^nk+(n+1)^2.$ Show this equality and you are done.
A: You could write $$\left( \sum_{k=1}^{n+1} k \right)^2 = \left[ \left(\sum_{k=1}^n k \right)+ (n+1) \right]^2 = \left(\sum_{k=1}^n k \right)^2 + 2(n+1) \left( \sum_{k=1}^n k\right)  + (n+1)^2$$
The induction step assumes that the first term on the right is just $\displaystyle \sum_{k=1}^n k^3$. You need to show that the rest of the terms add up to $(n+1)^3$.
A: I'd do this is using induction to prove that $\sum_{k=1}^n k = \frac{n}{2}(n+1)$ and $\sum_{k=1}^n k^3 = \frac{n^2}{4}(n+1)^2$; these are straightforward induction proofs. 
Then $$\left[\frac{n}{2}(n+1)\right]^2 = \frac{n^2}{4}(n+1)^2 \iff \left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^n k^3$$
A: I know you have to use induction, so this doesn't help you, but below is a cool proof by picture:

