You are proposing that $n^m > m^n \iff n > m$. However, there are many examples where this is not entirely true.

If $n = 2 \land m = 3 \implies n^m < m^n : n < m$

If $n = 2 \land m = 4 \implies n^m = m^n : n < m$

And obviously if $n = 1 \land m > 1 \implies n^m < m^n : n < m$

But perhaps what you are trying to say is that:

If $n > m \implies n^m > m^n$ because it seems like $n < m$ in these contradicting examples above. I mean, why do these seem to be the only contradicting examples? With the examples above, we know that $n \neq m \neq 0 \lor 1 \because n^m < m^n$. So moving from $1$ to $2$ where $n = 2$ and $m > 2$, we find a little shift in the equality signs.

For the first example, $n^m < m^n$

For the second example, $n^m = m^n$

And it seems that if $m > 2^2 = 4$ then your theory is true where $n^m > m^n$. And it seems like the reason your theory looks true on the condition that $m > 2^2$ is because we must find the first $n^m \lor m^n : n \land m > 1$ (because $1 > 0$ which is obviously $2^2$.

In summary, the theory is not that "if $n^m > m^n$ then $n > m$" but is instead that:

$$\text{if} \qquad n > m \implies n^m > m^n : n \land m \in \mathbb{W}$$

(since $\mathbb{W}$ is the set of all numbers $\ge 0$ aka "Whole Numbers")

You are proposing that $n^m > m^n \iff n > m$. However, there are many examples where this is not entirely true.

If $n = 2 \land m = 3 \implies n^m < m^n : n < m$

If $n = 2 \land m = 4 \implies n^m = m^n : n < m$

And obviously if $n = 1 \land m > 1 \implies n^m < m^n : n < m$

But perhaps what you are trying to say is that:

If $n > m \implies n^m > m^n$ because it seems like $n < m$ in these contradicting examples above. I mean, why do these seem to be the only contradicting examples? With the examples above, we know that $n \neq m \neq 0 \lor 1 \because n^m < m^n$. So moving from $1$ to $2$ where $n = 2$ and $m > 2$, we find a little shift in the equality signs.

For the first example, $n^m < m^n$

For the second example, $n^m = m^n$

And it seems that if $m > 2^2 = 4$ then your theory is true where $n^m > m^n$. And it seems like the reason your theory looks true on the condition that $m > 2^2$ is because we must find the first $n^m \lor m^n : n \land m > 1$ (because $1 > 0$ which is obviously $2^2$).

In summary, the theory is not that "if $n^m > m^n$ then $n > m$" but is instead that:

$$\text{if} \qquad n > m \implies n^m > m^n : n \land m \in \mathbb{W}$$

(since $\mathbb{W}$ is the set of all numbers $\ge 0$ aka "Whole Numbers")

You are proposing that $n^m > m^n \iff n > m$. However, there are many examples where this is not entirely true.

If $n = 2 \land m = 3 \implies n^m < m^n : n < m$.

If $n = 2 \land m = 4 \implies n^m = m^n : n < m$.

And obviously if $n = 1 \land m > 1 \implies n^m < m^n : n < m$.

But perhaps what you are trying to say is that:

If $n > m \implies n^m > m^n$ because it seems like $n < m$ in these contradicting examples above. I mean, why do these seem to be the only contradicting examples? With the examples above, we know that $n \neq m \neq 0 \lor 1 \because n^m < m^n$. So moving from $1$ to $2$ where $n = 2$ and $m > 2$, we find a little shift in the equality signs.

For the first example, $n^m < m^n$.

For the second example, $n^m = m^n$

And it seems that if $m > 2^2 = 4$ then your theory is true where $n^m > m^n$. And it seems like the reason your theory looks true on the condition that $m > 2^2$ is because we must find the first $n^m \lor m^n : n \land m > 1$ (because $1 > 0$ which is obviously $2^2$.

In summary, the theory is not that "if $n^m > m^n$ then $n > m$" but is instead that:

$$\text{if} \qquad n > m \implies n^m > m^n : n \land m \in \mathbb{W}$$

(since $\mathbb{W}$ is the set of all numbers $\ge 0$ aka "Whole Numbers")

You are proposing that $n^m > m^n \iff n > m$. However, there are many examples where this is not entirely true.

If $n = 2 \land m = 3 \implies n^m < m^n : n < m$

If $n = 2 \land m = 4 \implies n^m = m^n : n < m$

And obviously if $n = 1 \land m > 1 \implies n^m < m^n : n < m$

But perhaps what you are trying to say is that:

If $n > m \implies n^m > m^n$ because it seems like $n < m$ in these contradicting examples above. I mean, why do these seem to be the only contradicting examples? With the examples above, we know that $n \neq m \neq 0 \lor 1 \because n^m < m^n$. So moving from $1$ to $2$ where $n = 2$ and $m > 2$, we find a little shift in the equality signs.

For the first example, $n^m < m^n$

For the second example, $n^m = m^n$

And it seems that if $m > 2^2 = 4$ then your theory is true where $n^m > m^n$. And it seems like the reason your theory looks true on the condition that $m > 2^2$ is because we must find the first $n^m \lor m^n : n \land m > 1$ (because $1 > 0$ which is obviously $2^2$.

In summary, the theory is not that "if $n^m > m^n$ then $n > m$" but is instead that:

$$\text{if} \qquad n > m \implies n^m > m^n : n \land m \in \mathbb{W}$$

(since $\mathbb{W}$ is the set of all numbers $\ge 0$ aka "Whole Numbers")