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Bronya from Tattingstone and Alistair from Histon and Impington Infant School both drew a table to show the number of grains on each square (as suggested in the clue) to answer this problem. Here is Alistair's table:

Square Grains
1 1
2 2
3 4
4 8
5
16
6 32
7 64
8 128

Bronya carried on her table up to square 15. She explained that she saw the number of grains of rice were doubling each time and that these numbers were also powers of two. She wrote these powers down the side of the table like this:

Square Grains Powers of Two
1 1  
2 2  
3 4 $2^2$
4 8 $2^3$
5
16
$2^4$
6 32 $2^5$
7 64 $2^6$
8 128 $2^7$

Bronya then went on to say:

I noticed that the power of 2 was one less than the number of the square. So the number of grains of rice on the nth square would be $2^{n-1}$

Alistair then drew another table to show the number of grains altogether:

Square Total Grains
1 1
2 3
3 7
4 15
5 31
6 63
7 127

He then continued the table using approximate answers:

Square Total Grains
$7$ $12 \times10$
$8$ $25 \times10$
$9$ $51 \times10$
$10$ $10 \times 10^2$
$11$ $20 \times10^2$
$12$ $40 \times 10^2$
$13$ $80 \times 10^2$
$14$ $16 \times 10^3$
$15$ $32 \times 10^3$
$16$ $64 \times 10^3$
$17$ $12 \times 10^4$

Then Alistair said he noticed that square number 17 was 1000 times square number 7. So he took square 14 (which is equal $16 \times 10^3$) and multiplied it by 1000 to get an estimate for square 24, and so on... He continued the table in this way and arrived at an answer of $16 \times10^{18}$ by the 64th square. In other words approximately 16 000 000 000 000 000 000 grains of rice!

Both Alistair and Bronya noticed that the total number of grains up to and including a particular square was one less than the number of grains on the next square. They both wrote this down as a general rule using $n$:
$$(2^{n-1} \times2) -1$$ or more simply
$$2^n-1$$
 
As Bronya indicates, this would mean that the exact number of grains of rice on the entire board would be:
$$(2^{63} \times 2)-1$$ or
$$2^{64}-1$$

Fantastic work! Thank you to Bronya and Alistair for such logically reasoned solutions.

Logan, Gautam and Yash from Kohia Terrace School in New Zealand got the same formula, but also showed how to calculate the actual number of grains of rice:

The last question asks you how many grains of rice are on the entire board. First you have to figure out how to get there. Well if we look at the first three squares we find 1 grain + 2 grains + 4 grains = 7 grains of rice. We notice a doubling pattern so we can use the squares of two. There are 3 squares so far so the answer is $2^3 -1$ which $= 7$ grains. To figure out how many grains of rice are in the last square you need to figure out what $2 ^{63}$ is. And to find the total number of grains of rice you have to do $2^{n+1} - 1.$
$n$ is $63$ so the equation becomes: $2^{64}-1.$ You can double $2^{64}$ for ages but that takes a long time and there is a faster strategy. 

If you square $2$ you get $4$ which is the same as $4^1.$ So if we want to figure out $2^{64}$ we have to square the number that is getting powered (the $2$) and half the power number (the $64$).

If we do this we get $4^{32}$.

Again square $4$ to get $16$ then get $32 \div 2$ which $= 16.$ So $16^{16}$ is still the same question.

$16$ squared $= 256$  and $16\div2 = 8$ so again $256^8$  is still the same. 

$256$ squared is $65536$ and $8\div2 =4$ so $65536^ 4$ still has the same answer.

$65536$ squared $=4294967296$

So $4294967296^2$ is still the same as $2^{64}$

So if square one last time, we get $18246494073709551616$

Finally to finish the equation we have to minus 1 and I am happy to say, Sissa will get a total of $18246494073709551615$ grains of rice.


Extra for Experts

As well as figuring out the numerical answer we figured out the answer in Binary.

In Binary if you start with $1$ and you add one zero to the end: $10$, you get the number $2$ which is also $2^1.$ If you have 0b $100$, the number is $4.$ There are 2 zeros at the end of the number and $4$ is $2^2.$ If you keep checking it always works. To start we need $2^{64}.$ We get $1$ and 64 zeroes which is 0b $10000000000\ 0000000000\ 0000000000\ 0000000000\ 0000000000\ 0000000000\ 0000.$ But that is still not the right answer, We have to minus $1.$ The right answer is 0b $1111111111\ 1111111111\ 1111111111\ 1111111111\ 1111111111\ 1111111111\ 1111.$  Note: We have put the binary numbers in sets of 10 to make it easier to read.