Is 128 the only multi-digit power of 2 such that each of its digits is also a power of 2? The number $128$ can be written as $2^n$ with integer $n$, and so can its every individual digit. Is this the only number with this property, apart from the one-digit numbers  $1$, $2$, $4$ and $8$? 
I have checked a lot, but I don't know how to prove or disprove it. 
 A: This seems to be an open question.  See OEIS sequence A130693 and references there.
A: No value less than 2^30,000 bigger than 128 follows this pattern, as I have found through this tool:
http://www.michalpaszkiewicz.co.uk/maths/series/powers-of-two.html
You can use this tool to try and find such a value, but it would probably be easier to prove no such number exists using number theory.
A: Every such number is of the form of a sum of f(n,m) = 2^n*10^m where m is an integer and n is 0,1,2,3. You can gain insight into this problem by writing this problem out in binary. Every power of two is a 1 followed by zeros, so 1, 10, 100, 1000 is 1,2,4,8 etc. So multiplying a binary by 2 adds a zero to the right. So lets look at powers of ten:
1
1010
1100100
1111101000
10011100010000
The key point to note is that they are followed by more and more trailing zeros. Lets assume this is true. I knows its true up to 10^22 or so, since its used in algorithms to do fast floating point arithmetic, see e.g. here: http://www.exploringbinary.com/why-powers-of-ten-up-to-10-to-the-22-are-exact-as-doubles/ .
EDIT - This is obviously true, since every power of ten is a power of two since 10 = 2*5. Thus ever power of 10^n must have exactly n trailing zeros. (Since powers of 5 are odd they must end in one.)
I can add up to three zeros (multiply by 8), but i cannot exclude any power of ten. So I have to sum these. Clearly I have to eliminate all the 1's. So to convert 100 to the next power of two I take 1100100 and clearly need to add 0011100 i.e. fill in all the zeros with 1's and put a final one in to make them carry like dominoes. This is similar to making a power of ten, work out what you add to make the number nines followed by zeros then add one to which ever column has the last non zero digit. 
so can I make 11100 out of ten and one? yes, 10*2 = 10100 and 1*8 = 1000. So 128 is a perfect power.
Lets look at eliminating the 1s in the last two columns. Clearly all numbers of 100 or greater have no ones so don't matter, so I can have 12. I can also have 48. So any such number must end in 12, 28 or 48. Next try eliminating all the 1's in the last three columns. 128 = 10000 will do the job. So will 112 and 248, but thats it*. Lets follow the 112 chain. That will give us 2112 4112 8112 with 5 trailing zeros.
If we move to six trailing zeros we get 22112 42114 82112 14112 and 18112*. There are, around 1000 numbers with six trailing zeros under 10000. So if there is no deep structure, so we see that this form is extremely restrictive in terms of the sets that you can build. 
Its even conceivable that an exhaustive search along these lines would lead to all the branches terminating - i.e. that you couldn't eliminate the last one's any further. In that case you would have proved the statement. 
*I offer no guarantees that my inspection was exhaustive!
