I kind of stumbled across this pattern recently and cant make any sense of it. Attached is my work explaining what I did. If there is already an answer to this please post the link. Thanks!
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You've got a couple of errors on $n=3$, namely, $34-19\neq18$ and $61-34\neq24$. – barak manos Nov 01 '16 at 04:36
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Right, 34 should be 37, i copied my messy work quickly to a clean page. – Isaac Bowser Nov 01 '16 at 04:39
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http://math.stackexchange.com/questions/1878810/formula-for-1k2k3k-nk-for-n-k-in-mathbbn/1878949#187894 may be of interest. – Ahmed S. Attaalla Nov 01 '16 at 14:27
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Hint. If you form a row $$p(1)\quad p(2)\quad p(3)\quad\cdots$$ from any polynomial $$p(x)=ax^n+\cdots\ ,$$ then the next row will be $$q(1)\quad q(2)\quad q(3)\quad\cdots$$ where $$q(x)=nax^{n-1}+\cdots\ .$$
David
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It's more natural to start the sequence at $0$: for example,
$$0 \quad 1 \quad 4 \quad 9$$ $$1 \quad 3 \quad 5$$ $$2 \quad 2$$ $$0$$
from which you can read off the leftmost side that the top (the sequence $n^2$, $n \in \mathbb{N}$) is actually $$\color{red}{0} \cdot \binom{n}{0} + \color{red}{1} \cdot \binom{n}{1} + \color{red}{2} \cdot \binom{n}{2} = n + 2 \cdot \frac{n(n-1)}{2}.$$
This idea goes by the name Binomial transform.
The fact that you will end up with the last coefficient $k!$ when studying the sequence $1^k,2^k,3^k$ is because the coefficient of $n^k$ in $\binom{n}{k}$ itself is exactly $1/k!$; and you need the result to be $1 \cdot n^k.$