In its usual form, the Eckmann–Hilton argument shows that a monoid or group object in the category of monoids or groups is commutative. In other terms, if a set is equipped with two monoid structures, such that one is a homomorphism for the other, then the two structures coincide and the resulting monoid is commutative.
From the nPOV, we may want to think of the statement in this way:
Let $C$ be a 2-category and $x \in C$ an object. Write $Id_x$ for the identity morphism of $X$ and $End(Id_x)$ for the set of endo-2-morphisms on $X$. Then:
horizontal composition and vertical composition define the same monoid object structure on $End(Id_x)$;
this is an abelian monoid.
On the face of it, this is a special case of the general situation, although in fact every case is an example for appropriate $C$.
A more general version is this: If a set is equipped with two binary operations with identity elements, as long as they commute with each other in the sense that one is (with respect to the other) a homomorphism of sets with binary operations, then everything else follows:
This can also be internalised in any monoidal category.
A pasting diagram-proof of 1 is depicted in Cheng below. Here we prove the $6$-element general form in $Set$.
The basic equation that we have (that one operation $*$ is a homomorphism with respect to another operation $\circ$) is
In $End(Id_x)$, this is the exchange law.
We prove the list of results from above in order:
Simply read the basic equation backwards to see that $\circ$ is a homomorphism with respect to $*$.
Now if $1_*$ is the identity of $*$ and $1_\circ$ is the identity of $\circ$, we have
A similar argument proves the other half.
Then
so the identities are the same; we will now write this identity simply as $1$.
Now
so the operations are the same; we will write them both with concatenation.
Then
so this operation is commutative.
Finally,
so the operation is associative.
If you start with a monoid object in $Mon$, then only (4&5) need to be shown; the others are part of the hypothesis. This classic form of the Eckmann–Hilton argument may be combined into a single calculation:
where the desired results involve the first, middle, and last expressions.
A $2$-tuply monoidal $0$-category, if defined as a pointed simply connected bicategory, is also the same as an abelian monoid.
A $2$-tuply monoidal $1$-category, if defined as a pointed simply connected tricategory, is the same as a braided monoidal category.
Every homotopy group $\pi_n$ for $n \geq 2$ is abelian.
The beautiful and powerful Eckmann-Hilton argument is due to Beno Eckmann and Peter Hilton.
An expositions of the argument is given here:
The diagram proof is displayed here
and an animation of it is here
For higher analogues see within the discussion of commutative algebraic monads at: