If $n>m>0, n, m \in \Bbb{N}$, is $n^m-m^n$ positive or negative? (Explain with cases if needed.)

Question

If $n>m>0, n, m \in \Bbb{N}$, is $n^m-m^n$ positive or negative? (Explain with cases if needed.)

First, for finding patterns, I've put some small integers to $(n, m).$

\begin{align} (2, 1): \; & 2^1-1^2=1>0 \\ (3, 2): \; & 3^2-2^3=1>0 \\ (4, 3): \; & 4^3-3^4=-17<0 \\ (5, 4): \; & 5^4-4^5=-399<0 \\ \end{align}

...and gained nothing.

Also, I've found some trivial theorem about this, like:

---

Thm 1. $n^1-1^n>0$ for all $n>1$.

No need to prove it.

---

Thm 2. $n^2-2^n<0$ for all $n>4$.

Proof:

Let's draw a graph for this.

The red one is a graph of $y=x^2$, and the blue one is a graph of $y=2^x$.

As you can see, $y=2^x$ increases more exponentially then $y=x^2$.(Also a trivial thing without graph)

The intersection of two graphs in positive reals is $(2, 4), (4, 16)$.

Therefore, $n^2-2^n<0$ for all $n>4$.

---

I think there is a particular integer $N$ which satisfies, $n^N-N^n<0$ for $\forall n>N$.

But I can't do anything more...