The aim is to find the nth derivative of the given function using a known series from $e^x$.
$$ e^x = \sum\_{n \to 0}^ \infty { \frac {x^n} {n!} } $$
and now to find 

$$ f(x) = -x^5 {e^{x^3-1}}$$
$$ f(x) = -x^5 \frac {e^{x^3}} {e} $$
and so after using the known series to find this series above we get...
$${ \frac {1}{e}} \sum\_{n \to 0}^{ \infty } (-1)^n \frac {x^{3n+5}}{n!}$$

and we have to find the $f^{(100)}(0)$ I did it like $ 3n + 5 = 100 \to 3n = 95 \to n = \frac {95}{3} \notin \Bbb N$
and also find $f^{(101)}(0)$ = 32 (similar method as above), and for 32 we get $$(-n)^{32}. \frac {101!} {32!}$$

Could anyone please spot my mistake and show me the proper method to do so? I'm kinda new to infinite series.