The question looks like this:

1000^{1001}(?) 1001^{1000}We want to find the (?) operator.

I scribbled that a moment ago. I'm using *logarithm*, and captured the typed-steps as an image.

Let's start

Hang on for a moment...

In conclusion

Without using calculator,

1000^{1001}is larger than1001^{1000}

`1000`

^{1001}> 1001^{1000}

Closure

As I observed, that pattern starts from integer greater than 2 (to any large positive integer).

I haven't tinkered the positive "infinity". I suppose it will be the *same*.

It's *like* imagination, the infinity concept.

And for number with, like, floating point, you can do your own scribbling.

Anyway, if we use 1 (in that pattern above):

1^{2}(?) 2^{1}

1 (?) 2

1 < 2

If we use 2:

2^{3}(?) 3^{2}

8 (?) 9

8 < 9

The next one, 3:

3^{4}(?) 4^{3}

81 (?) 64

81 > 64

Next, 4:

4^{5}(?) 5^{4}

1024 (?) 625

1024 > 625

And so on...

This technique will surely result different outcome for integer less than or equal to 2. Especially on the Let's assume it's 1 step (rounding down step).

Wolfram and Wikipedia

This is the link on

**Wolfram|Alpha**about that particular question.This is the the form equivalency and decimal approximation of

*log*on Wolfram|Alpha._{1000}(1001)This is the list of

*logarithmic identities*on Wikipedia.
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