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To do a proof by strong induction, we follow 3 steps:

Statement: Begin with a precise statement of the formula to be proven.

The Basis Case: State the number $k$ where you're starting your induction, and the formula $f(k)$ to be proven, and then prove it.

The Induction Case: State the inductive hypothesis $f(n)$, and the formula $f(n+1)$ to be proven. Prove the result, clearly indicating when the inductive hypothesis is used.

We have to give a proof for the following theorem:

2) Formalize the theorem: $$\sum_{i=1}^{n}i^k = \frac{1}{k+1}\cdot n^{k+1}$$ by proving by induction $\nabla n^{(k)}=kn^{(k−1)}$ for all natural numbers $k\geq 1$. It may help to first prove the product rule for differentiation: $$\nabla f(n)\cdot g(n)=f(n)\nabla g(n)+g(n−1)\nabla f(n) .$$

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    Note that $\sum\limits_{i=1}^{n+1} a_i = a_{n+1}+\sum\limits_{i=1}^n a_i$... i.e. you can "*break off a piece*" of the sum to be computed separately. – 2017-01-21
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    While $\Delta n^{(k)} = kn^{(k-1)}$, $\nabla n^{(k)} = k(n-1)^{(k-1)}$... – 2017-01-21
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    Also, the "theorem to be formalized" is, I suspect, $\sum_{i=0}^{n-1} i^{(k)} = n^{(k+1)}/(k+1)$. – 2017-01-22
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    Have you tried plugging in a few numbers? You may notice that $n^{k+1}/(k+1)$ is often not an integer, while the left-hand side always is. (And so is $n^{(k+1)}/(k+1)$.) – 2017-01-25
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    I'm voting to close this question as off-topic because the statement is wrong, no one has been interested enough to post an answer, and OP hasn't been interested enough to engage with the comments. – 2018-10-02

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