$\sum_{n=1}^{1337}(n!)^4$

What can you say about the units digit of
$n!$
for$n\ge5$
? 
What about the units digits for
$n < 5$
? 
It’s sufficient to take only the units digit when multiplying or raising the number to any power.

All factorials greater that
$5!$
have both the factors$5$
and$2$
, hence the units digit equal to 0. This means that we only need to worry about$n \in \{1,2,3,4\}$
. It’s sufficient to take only the units digit when multiplying or raising to any power.We have:
$1^4\rightarrow1$
$(2!)^4=2^4\rightarrow6$
$(3!)^4=6^4\rightarrow6$
$(4!)^4=24^4\rightarrow4^4\rightarrow6$
The units digit for the sum is therefore the units digit of
$1+6+6+6$
, i.e.$9$
.