Evaluating 8th derivative of $(e^x-1)^6$ at $x=0$

Question

8 distinct objects are distributed into 7 distinct boxes. Find the number of ways in which these objects can be distributed to exactly 6 boxes.

I have solved this question quite easily using method of division and distribution ${7 \choose 6}(\frac{8! \times 6!}{5!\times 3!} + \frac{8! \times 6!}{4! \times (2!)^2 \times 2!})$, And also using principle of inclusion and exclusion ${7 \choose 6}(\sum_{r=1}^{6} {6 \choose r} r^8 (-1)^r)$ so please do not solve this question.

The question in the title:

Evaluating 8th derivative of $(e^x-1)^6$ at $x=0$

arose as I was thinking of evaluating the expression (I have excluded the ${7 \choose 6}$ term as that doesn't affect my question) obtained in the method using principle of inclusion and exclusion using generating functions, as the calculations were getting a bit too messy.

I'm also unable to evaluate this expression easily. Is there a way we can evaluate this expression easily? (Preferably without any calculator).

I deduced that the $(e^x-1)$ factor must be differentiated 6 times, or else at x=0 it would become 0. However I'm not able to proceed; the additional $e^x$ factors from chain rule of is making it still tedious for me.

I have verified using Wolfram alpha that all the three expressions give the same answer $266*7!$.

Answer 1

We need to find the $x^8$ coefficient of $(e^x-1)^6$. This is the $x^2$ coefficient of $[(e^x-1)/x]^6$. But $$\frac{e^x-1}x=1+\frac x2+\frac{x^2}6+\text{higher terms}.$$ So we need to find the $x^2$ term of $$\left(1+\frac x2+\frac{x^2}6\right)^6.$$ But $$(1+ax+bx^2)^n=1+nax+\left(nb+\binom n2a^2\right)x^2+\text{higher terms}.$$ So $$\left(1+\frac x2+\frac{x^2}6\right)^6 =1+3x+\left(1+\frac{15}4\right)x^2+\text{higher terms}.$$ So the $x^8$-coefficient of $(e^x-1)^6$ is $19/4$.

So your eighth derivative is $8!$ times this, which is $38\times 7!=266\times 6!$.

Answer 2

Use series. You know the series for $e^x$, and can get the expansion:

$\begin{align*} (e^x - 1)^6 &= \sum_k (-1)^k \binom{6}{k} e^{k x} \\ [x^8] (e^x - 1)^6 &= \sum_k (-1)^k \binom{6}{k} [x^8] e^{k x} \\ &= \sum_{1 \le k \le 6} (-1)^k \binom{6}{k} \frac{k^8}{8!} \\ &= \frac{1}{8!} \sum_{1 \le k \le 6} (-1)^k \binom{6}{k} k^8 \end{align*}$

A CAS gives $19 / 4$.