Property of powers of 2

Suppose 2^k=1a₁a₂....a_j
Here k=1+j...
Now folowing is the list of the 1st 2 digits of the coming powers...

2 or 3
4 or 5 6 or 7
8 or 1.... 1... or 1....
1....

Easy to see that these nos contain j+2 digits with leading digit 1....
And it is also quite obvious that there shall be exactly 1 such power, for example suppose there exists another power, 2^l, where l=k+a.....then it is easy to visualise the leading digit does not remain 1 anymore, thus proved, that if it is true for k=j+1, it's true for k'=j+2 as well, j=1 satisfies the statement (2⁴=16), thus proved.

let p(k) denote a power of 2 with k digits with 1 as the leading digit
10¹<p(2)<2(10¹)
10²<p(3)<2(10²)
suppose 10^k-1<p(k)<2(10^k-1) {if this is proved, uniqueness is self explanatory}
multiplying throughout by 2³
(8)10^k-1<(2³)p(k)<16(10^k-1)
case 1
10(10^k)<(2³)p(k)<16(10^k-1)
so (2³)p(k) = p(k+1)
case 2
(8)10^k-1<(2³)p(k)<(10)(10^k-1)
multiplying throughout by 2
(16)10^k-1<(2⁴)p(k)<(20)(10^k-1)
so (2⁴)p(k)=p(k+1)

now it can be seen either of case 1 or 2 is true but never both
uniqueness of the result is as below

please omit the last line 'uniqueness of the result is as below'

sir is my solution correct?