Expression for power of a natural number in terms of binomial coefficients

The coefficients $6,6$ and $1$ in that expansion are closely related to Stirling numbers of the second kind. There is an identity

$$x^n=\sum_k{n\brace k}x^{\underline k}=\sum_k{n\brace k}x(x-1)(x-2)\dots(x-k+1)\;,$$

which in the case $x=m\in\Bbb N$ becomes

$$m^n=\sum_k{n\brace k}\frac{m!}{(m-k)!}=\sum_k{n\brace k}k!\binom{m}k\;,$$

so that the coefficients are the numbers $\displaystyle k!{n\brace k}$. I didn’t bother to specify the range of $k$, because it’s implicitly determined by the Stirling numbers themselves: for $n>0$, the only values of $k$ for which $\displaystyle{n\brace k}\ne0$ are $k=1,\dots,n$.

A good place to learn about such things is the excellent book Concrete Mathematics, by Graham, Knuth, & Patashnik.

See the Wikipedia under Binomial coefficients as a basis for the space of polynomials.

In your particular case, and using the notation of the Wikipedia article, from which I am quoting formulas, $$ t^{d} = \sum_{k=0}^d a_k \binom{t}{k}, $$ where $$ a_k = \sum_{i=0}^k (-1)^{k-i} \binom{k}{i} i^d. $$