Sextuple Integral

Looking at a $\frac{6!}{n!}$ term, we integrate this $(6-n)$ times. When integrating we divide by the power of next power of $x$ produce, so in these $(6-n)$ integrations we end up dividing by $(6-n)!$ . Hence, $\iiint\limits_{6-n} \int_0^1{\frac{6!}{n!}} = \frac{6!}{n!(6-n)!} = \binom{6}{n}$ .

Our result is hence $\binom{6}{5} + \binom{6}{4} + \binom{6}{3} + \binom{6}{2} + \binom{6}{1} + \binom{6}{0} = \left (\sum_{c=0}^6 \binom{6}{c} \right ) - \binom{6}{6} = 2^6 - 1 = 63$

$\int_0^1 \left(\int_0^f \left(\int_0^e \left(\int_0^d \left(\int_0^c \left(\int_0^b \frac{6!}{0!} \, da+\frac{6!}{1!}\right) \, db+\frac{6!}{2!}\right) \, dc+\frac{6!}{3!}\right) \, dd+\frac{6!}{4!}\right) \, de+\frac{6!}{5!}\right) \, df \Rightarrow 63$

$720 b,\,360 c^2+720 c,\,120 d^3+360 d^2+360 d,\,30 e^4+120 e^3+180 e^2+120 e,\,f^6+6 f^5+15 f^4+20 f^3+15 f^2+6 f$