But then I started thinking harder - what if our calculator is a very basic four-function calculator?  Effectively making this a pencil and paper problem.

The big number is less than 10^69, so we know the root is at most three digits, and specifically we can say it is less than 1001.

Let's start by calculating the big number mod 1001.  This is much like the mod 11 rule but just operating on three-digit blocks.  Then we calculate:

-70+843-029+071-054+204-781+573-025+995-798+769-907+739-696+815-288+329-699+328 = 1319

and then 1319 mod 1001 = 318.

Now 1001 = 7*11*13, so let's break 318 mod 1001 down:

318 = 3 mod 7

318 = 10 mod 11

318 = 6 mod 13

Then we have the following system of congruences:

Next we raise each side to a power so the exponent of N is one more than a multiple of one less than the mod base.

N^(6*19+1) = (N^23)^5 = 3^5 mod 7

N^(10*16+1) = (N^23)^7 = 10^7 mod 11

N^(12*21+1) = (N^23)^11 = 6^11 mod 13

Now we are set to apply Fermat's Little Theorem to the left side of each equation, which will reduce to N in all three cases.  Then

N = 3^5 = 243 = 5 mod 7

N = 10^7 = 10*100^3 mod 11 = 10*1^3 = 10 mod 11

N = 6^11 = 36*216^3 = 10*8^3 = 5120 = 11 mod 13

So now we are down to a simple system of congruences: N=5 mod 7, N=10 mod 11, and N=11 mod 13.

One more thing to find are the modular inverses: 

(11*13)^-1 mod 7, (7*13)^-1 mod 11 and (7*11)^-1 mod 13.

11*13*5=102*7+1, so (11*13)^-1 = 5 mod 7

7*13*4=33*11+1, so (7*13)^-1 = 4 mod 11

7*11*12=71*13+1, so (7*11)^-1 = 12 mod 13

Now our final calculation is to apply the Chinese remainder theorem.  Then N = 5*5*(11*13) + 10*4*(7*13) + 11*12*(7*11) = 3575+3640+10161 = 17379 = 362 mod 1001.

That is our answer 362.

This is a LOT of work.  Quite doable, though very tedious, with only a very basic calculator. But to make this work we needed some theorems from basic Number Theory, which are unlikely to be taught at the high school level, whereas logarithms and exponents are part of a typical high school algebra course.