Regular Expression for replacing Leading Spaces

I don't claim to be any expert on Regular Expressions and even after reading CD's introduction to Reg Exp, I still can't figure out this one which I'm sure must be very basic.
I want to replace all the leading spaces in a string with "." chrs. I could do this using the common replace/substr/instr functions, but I reckoned it would be possible in a single regular regexp_replace call.
So far I've got this...
SQL> select regexp_replace('      FRED BLOGS    WAS HERE    ', '^([:space:])*', '.')
  2  as result
  3  from dual;
RESULT
.      FRED BLOGS    WAS HERE
SQL>Which is replacing the start of line with a "." and not the spaces.
But I want my result to be:-
RESULT
......FRED BLOGS    WAS HERE
SQL>Cheers

That was very good solution .
Can you explain me the significance of "| " in the code, other things I could trace out.
I try to run the code with the 2 cases
when I give a space after | symbol it prints the * many times
SQL> SELECT col1, REGEXP_REPLACE(col1, ' ([^ ]+.*)|','*\1')
2 FROM (SELECT ' FRED BLOGS WAS HERE ' col1 FROM dual );
COL1
REGEXP_REPLACE(COL1,'([^]+.*)|','*\1')
FRED BLOGS WAS HERE
*FRED BLOGS    WAS HERE
SQL> SELECT col1, REGEXP_REPLACE(col1, ' ([^ ]+.*)| ','*\1')
2 FROM (SELECT ' FRED BLOGS WAS HERE ' col1 FROM dual );
COL1
REGEXP_REPLACE(COL1,'([^]+.*)|','*\1')
FRED BLOGS WAS HERE
******FRED BLOGS WAS HERE

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