You asked WA for the result with 100 digits of precision. It probably is 1 to that precision.

There is no need to guess. First of all:

    (1 + 1/n)^n = e^(log(1 + 1/n) * n)

Now from the Taylor series approximation:

    log(1 + t) = t - t^2/2 + t^3/3 - t^4/4 + ...

Substitute that in and we get:

    e^(log(1 + 1/n) * n)
        = e^((1/n - 1/(2 n^2) + 1/(3 n^3) - ...) n)
        = e^(1 - 1/(2n) + 1/(3n^2) - ...)

And now you can apply the tangent line approximation to find that this is:

    e - e/(2n) + O(1/n^2)

So if n = 10^N, then the first error should be around the N'th digit.

I don't think it is. Try N[((1+10^-n)^(10^n))/e, n] for smaller values of n. At first it approaches e faster than the precision can "catch up", and up until around n=50 it will result in 1. But at 10^60 and above it will return 1.0000000…