The Powers of $32$ !

If we rewrite the numbers in the base of 32, we get the consecutive binary natural numbers.

$(35467067393)_{10}=(11101001)_{32}$ ,So the answer is $(11101001)_2=(233)_{10}$

The distinctness of the powers and a specific base instantly forces one to think about base conversions. We try base $32$ and rewrite the sequence:

$1,10,11,100,101,...$

It's the list of consecutive binary integers (it's obvious since the expansion of terms don't have any additional coefficients). The rest is straightforward.

$35467067393=11101001_{32}$

$11101001_2=233$

$\therefore ~~~ N=\fbox{233}$

Let's write this sequence in a different way.. we will convert each number to base 32:

$[1, 10, 11, 100, 101, 110, 111, 1000, 1001, ...]$

Those familiar with binary should recognize a very clear pattern. These are the consecutive natural numbers in base 2!

After some calculations (find the greatest $n$ such that $32^n < 35467067393$ , subtract and repeat), it can be found that $35467067393 \equiv 11101001$ when represented in base 32. To calculate this number's place in the sequence, we simply convert $11101001$ to decimal, as if it were binary.

$1 + 8 + 32 + 64 + 128 = \boxed{233}$