Help in Independence Probability Again

I need help again -__-

A fair die is rolled and then n fair coins are tossed, n being the number shown on the die. What is the probability that no heads are shown?

Answer key says it's 63 / 384.

Probability of finding any number when a fair die is rolled = 1/6
Pr. of getting a head when a fair coin is tossed = 1/2
Number
on the die...P ... P(no heads) ....
1 ...........1/6 ... 1/2.................... 1/6*1/2 = 1/12
2 ...........1/6 ... (1/2)^2 .............. 1/6*1/4 = 1/24
3 ...........1/6 ... (1/2)^3 .............. 1/6*1/8 = 1/48
4 ...........1/6 ... (1/2)^4 .............. 1/6*1/16 = 1/96
5 ..........1/6 .....(1/2)^5 .............. 1/6*1/32 = 1/192
6 ......... 1/6 .... (1/2)^6 ............... 1/6*1/64 = 1/384
......................................… (32+16+8+4+2+1)/384 = 63/384

Now its like this
Probability of the number 1 coming on die is 1/6
If this comes, then only one coin is tossed which means probability of HEAD not coming is 1/2.
Total probability is 1/6*1/2 = 1/12.
Next Independent event is 2 coming on die.
Its probability = 1/6.
Now 2 coins are tossed. So the probability of HEAD not coming on any of the coins is 1/4
(TAIL COMES ON 1st COIN * TAIL COMES ON 2nd COIN) = (1/2*1/2)
Thus probability of this event is 1/4*1/6
Similarly, for number 3 on die, we get
1/6 * 1/2*1/2*1/2
for 4
1/6 * 1/2*1/2*1/2*1/2
for 5
1/6 * 1/2*1/2*1/2*1/2*1/2
for 6
1/6 * 1/2*1/2*1/2*1/2*1/2*1/2
(Since 3 coins mean three tails, 4 means 4 etc)
Now we add all these probabilities (since these are independent events)
1/6*1/2 + 1/6*(1/2)*1/2 + .....
which amounts to your answer!

there is a 1/6 probability of rolling a 1, therefore chance is 0.50 to not see a heads
there is a 1/6 probability of rolling a 2, therefore chance is (0.50)^2 to not see a heads
there is a 1/6 probability of rolling a 3, therefore chance is (0.50)^3 to not see a heads