Sequence limit

(n!)^(1/n)
= exp{ (1/n) [ln(n!)] }
= exp{ (1/n) [ sum from 1 to n (ln k) ] }
= exp{ (1/n) [ sum from 1 to n (ln (k/n))] + n[ln(n)] }

when n tends to infinity ,

(n!)^(1/n) = exp{ lim n->oo (1/n) [ sum from 1 to n (ln (k/n))] + n[ln(n)] }

= exp{ lim n->oo (1/n)[n ln(n) ] } * exp { lim n->oo (1/n) [ sum from 1 to n (ln (k/n))] }

= ( lim n->oo n )* exp{ integrate ln(x) wrt x 0<= x <= 1 }
= ( lim n->oo n )* e(-1)
= oo

Actually, if we observe that
n! > (n/2)^(n/2), (just consider the second half of the product)
then it is obvious that the limit is infinity.

The limit does not exist
Take the logaritm of x_n we have (1/n)(ln1+ln2+...+lnn). It is larger than ∫1n x dx (from 1 to n) This integral equals nln n-n+1. Go back to the e-power and we'll get:
>n√(n!) > e^(nln n-n+1)
This shows that the sequence tends to infinity
Note: Since n√(n!) /n=1/e, it also implies that n√(n!) tends to infinity !