You can use

$o=sprintf "%10.4e",$o;
print "$o\n";
[download]

1.4100e-03
1.9881e-06
2.8032e-09
3.9525e-12
5.5730e-15
7.8579e-18
1.1080e-20
1.5623e-23
2.2028e-26
3.1059e-29
4.3793e-32
[download]

If you don't like it to be in scientific notation you can add a $o+=0; between the two lines to force the value back to numeric:

0.00141
0.0000019881
0.0000000028032
0.0000000000039525
0.000000000000005573
0.0000000000000000078579
0.00000000000000000001108
0.000000000000000000000015623
0.000000000000000000000000022028
0.000000000000000000000000000031059
0.000000000000000000000000000000043793
[download]

There are many ways to solve the problem, but here's a trick that may not leap straight to your finger tips:

use warnings;
use strict;
use bignum;

my $e = 0.00141;
my $o =1;
for (1 .. 10){
    "@{[$o *= $e]}" =~ /(0\.0*\d{0,5})/;
    print "$1\n";
}
[download]

Prints:

0.00141
0.0000019881
0.0000000028032
0.0000000000039525
0.0000000000000055730
0.0000000000000000078580
0.000000000000000000011079
0.000000000000000000000015622
0.000000000000000000000000022027
0.000000000000000000000000000031059
[download]

The @{[...]} trick is useful for interpolating the result of a chunk of code into a string. The rest should be pretty obvious, but note that if the regex doesn't match the print may generate some interesting results.

Update: bah, forgot about the rounding. Oh well, the interpolation trick is still worth knowing even if it doesn't really help solve the OP's problem.

The @{[...]} trick is useful for interpolating the result of a chunk of code into a string.

No doubt a nice trick... but in this particular case, the conversion to string would happen implicitly, i.e. you could also do

    ($o *= $e) =~ /(0\.0*\d{0,5})/;
[download]

Yet another way would be to use bignum's accuracy option (which also does take care of the rounding)

use bignum accuracy => 5;

my $e = 0.00141;
my $o = 1;
for (1 .. 10) {
    $o *= $e;
    print "$o\n";
}
__END__
0.0014100
0.0000019881
0.0000000028032
0.0000000000039525
0.0000000000000055730
0.0000000000000000078579
0.000000000000000000011080
0.000000000000000000000015623
0.000000000000000000000000022028
0.000000000000000000000000000031059
[download]

(see Math::BigInt and Math::BigFloat for details on rounding modes, accuracy, precision, etc.)

and than deal with that number when calculating the next number in the iteration (last number in the list)

This suggests to me that you actually want to use the rounded number in the next line of your iteration. Is this intentional?

A method that doesn't rely on interpreting the number in string context is as follows:

use warnings;
use strict;
use bignum;

sub trim
{
    my $number=$_[0];
    my $no_d_p=$_[1];  #number of decimal places
    die "$number is not a number" unless &isNumber($number); #check th
+at you've got the right input
    die "$no_d_p is not an integer" unless int($no_d_p)==$no_d_p;

    my $power=0;   
    for($power=0;$number<=(10**($no_d_p-1));$power++){
        $number*=10;
    }   
    $number=int($number);
    $number/=(10**$power);      
    return($number);
}

sub isNumber
{
   my $number=$_[0];
    if ($number !~ /^\-?\d+(?:\.\d+)?(?:[Ee][+-]?\d{1,3})?$/){
    return 0;
    } else {
    return 1;
    }
}

my $e = 0.00141;
my $o =1;
for(my $r=0; $r<=10; $r++){
        $o *=$e;
        $o=&trim($o,5);
        print "$o\n";
}


exit 0;
[download]

0.00141
0.0000019881
0.0000000028032
0.0000000000039525
0.000000000000005573
0.0000000000000000078579
0.000000000000000000011079
0.000000000000000000000015621
0.000000000000000000000000022025
0.000000000000000000000000000031055
0.000000000000000000000000000000043787
[download]

This obviously introduces rounding errors. is that what you want or have I mis-interpreted the question?