If you must do this, then the easiest way will be to just use the built in formating, and then just replace the e+05 or e-12 with whatever you'd rather have:

def sci_notation(number, sig_fig=2):
    ret_string = "{0:.{1:d}e}".format(number, sig_fig)
    a, b = ret_string.split("e")
    # remove leading "+" and strip leading zeros
    b = int(b)
    return a + " * 10^" + str(b)

print sci_notation(10000, sig_fig=4)
# 1.0000 * 10^4

from decimal import Decimal

'%.2E' % Decimal('40800000000.00000000000000')

# returns '4.08E+10'

In your '40800000000.00000000000000' there are many more significant zeros that have the same meaning as any other digit. That's why you have to tell explicitly where you want to stop.

If you want to remove all trailing zeros automatically, you can try:

def format_e(n):
    a = '%E' % n
    return a.split('E')[0].rstrip('0').rstrip('.') + 'E' + a.split('E')[1]

format_e(Decimal('40800000000.00000000000000'))
# '4.08E+10'

format_e(Decimal('40000000000.00000000000000'))
# '4E+10'

format_e(Decimal('40812300000.00000000000000'))
# '4.08123E+10'