That would probably match if not beat join, if you weren't stringifying the array twice.

Actually you did that previously and I carelessly cargo-culted it into my code.
However upon testing it's actually very slightly faster! I don't know why:

perl -MBenchmark=:all -MAlgorithm::Combinatorics=:all -wle'
for ("1".."4") {
my $t0 = Benchmark->new;
my $i  = variations_with_repetition(["a".."z"],$ARGV[0]); 
my @x; push @x, pack "(A*)*", @$_ while $_=$i->next;
my $t1 = Benchmark->new;
my $ts = timestr(timediff($t1,$t0));
print $ts }' 4
[download]

1 wallclock secs ( 0.77 usr +  0.01 sys =  0.78 CPU)
1 wallclock secs ( 0.77 usr +  0.01 sys =  0.78 CPU)
1 wallclock secs ( 0.75 usr +  0.00 sys =  0.75 CPU)
1 wallclock secs ( 0.75 usr +  0.01 sys =  0.76 CPU)

perl -MBenchmark=:all -MAlgorithm::Combinatorics=:all -wle'
for ("1".."4") {
my $t0 = Benchmark->new;
my $i  = variations_with_repetition(["a".."z"],$ARGV[0]); 
my @x; push @x, pack "A*", qq[@$_] while $_=$i->next;
my $t1 = Benchmark->new;
my $ts = timestr(timediff($t1,$t0));
print $ts }' 4
[download]

0 wallclock secs ( 0.74 usr +  0.01 sys =  0.75 CPU)
1 wallclock secs ( 0.72 usr +  0.01 sys =  0.73 CPU)
1 wallclock secs ( 0.73 usr +  0.00 sys =  0.73 CPU)
1 wallclock secs ( 0.74 usr +  0.01 sys =  0.75 CPU)

perl -MBenchmark=:all -MAlgorithm::Combinatorics=:all -wle'
for ("1".."4") {
my $t0 = Benchmark->new;
my $i  = variations_with_repetition(["a".."z"],$ARGV[0]); 
my @x; push @x, join "", @$_ while $_=$i->next;
my $t1 = Benchmark->new;
my $ts = timestr(timediff($t1,$t0));
print $ts }' 4
[download]

1 wallclock secs ( 0.65 usr +  0.01 sys =  0.66 CPU)
1 wallclock secs ( 0.64 usr +  0.00 sys =  0.64 CPU)
0 wallclock secs ( 0.63 usr +  0.00 sys =  0.63 CPU)
1 wallclock secs ( 0.64 usr +  0.01 sys =  0.65 CPU)

BTW, that is one seriously quick machine you're running. What is it?

MacBook Pro i7
3.1GHz (4.1GHz Turbo Boost) 16GB RAM 1TB SSD
Darwin 10.13.4