Prove that the sum of the terms of the $n$-th row of Euler's triangle is $n!$ I know that Euler's Triangle is related to the coefficients of the series
$$\sum_{n=0}^{\infty}n^ka^n,|a| <1$$
For example,
$$\sum_{n=0}^{\infty}na^n = \frac{a}{(1-a)^2}$$
$$\sum_{n=0}^{\infty}n^2a^n = \frac{a^2+a}{(1-a)^3}$$
$$\sum_{n=0}^{\infty}n^3a^n = \frac{a^3+4a^2+a}{(1-a)^4}$$
$$\sum_{n=0}^{\infty}n^4a^n = \frac{a^4+11a^3+11a^2+a}{(1-a)^5}$$
If we arrange the coefficients of the terms $\cdots a^4, a^3,a^2,a$, we get Euler's Triangle:

Note that $$ 1+1 = 2!$$
$$ 1+4+1=3!$$
$$1+11+11+1=4!$$
$$1+26+66+26+1 = 5!$$
It is known that the sum of the $n^{th}$ row is $n!$.
I am wondering if there is an actual proof for the general rule.
 A: Observe that, for each $r\in\mathbb{Z}_{\geq 0}$, we have $$\sum_{n=0}^\infty\,\binom{n}{r}\,a^n=\frac{a^r}{(1-a)^{r+1}}\text{ for all }a\in\mathbb{C}\text{ with }|a|<1\,.$$
Now, 
$$n^k=\sum_{r=0}^k\,t(k,r)\,\binom{n}{r}$$
for some integers $t(k,0)$, $t(k,1)$, $t(k,2)$, $\ldots$, $t(k,k)$.  In particular, $t(k,k)=k!$.  That is, for a complex number $a$ with $|a|<1$, we have
$$\sum_{n=0}^\infty\,\binom{n}{k}\,a^n=\sum_{n=0}^\infty\,\sum_{r=0}^k\,t(k,r)\,\binom{n}{r}\,a^n=\sum_{r=0}^k\,t(k,r)\,\sum_{n=0}^\infty\,\binom{n}{r}\,a^n\,.$$
Therefore,
$$\sum_{n=0}^\infty\,\binom{n}{k}\,a^n=\sum_{r=0}^k\,t(k,r)\,\left(\frac{a^r}{(1-a)^{r+1}}\right)=\frac{\sum\limits_{r=0}^k\,t(k,r)\,a^r\,(1-a)^{k-r}}{(1-a)^{k+1}}\,.$$
Hence, the sum of the coefficients of $a$ in the numerator of the expression above is simply
$$\lim_{a\to 1^-}\,(1-a)^{k+1}\,\sum_{n=0}^\infty\,\binom{n}{k}\,a^n=\lim_{a\to1^-}\,\sum\limits_{r=0}^k\,t(k,r)\,a^r\,(1-a)^{k-r}=t(k,k)=k!\,.$$
I would be interested to see a proof that the coefficient of $a^l$ in $f_k(a):=\sum\limits_{r=0}^k\,t(k,r)\,a^r\,(1-a)^{k-r}$  equals the coefficient of $a^{k+1-l}$ in $f_k(a)$ for every $l=1,2,\ldots,k$.
