Recall that we define the simplex category $\mathbf{\Delta}$ and define the category of simplicial sets $\mathbf{\Delta\mbox{-}Set}$ to have as objects, functors $\mathbf{\Delta}^\mathrm{op}\to\mathbf{Set}$, and as morphisms, natural transformations between these functors. We then define, for $n\in\mathbb{N}$, the simplicial n-simplex $\Delta^n$ s.t. $\Delta^n(i)=\mathrm{Hom}_\mathbf{\Delta}([i],[n])$ with the induced face and degeneracy maps.
Then we define the $(n,k)$-horn to be the minimal simplicial subset $\Lambda_k^n$ of $\Delta^n$ such that $$\Big(\delta^i:[n-1]\to[n]\Big)\in\Lambda_k^n(n-1)$$ for all $i\neq k$.
However, I'm having trouble making any concrete statements about $\Lambda_k^n$. How do we know that $\Lambda_k^n\ne\Delta^n$? How can it be shown that the non-degenerate simplices of $\Lambda_k^n$ do in fact form a horn, as the name suggests?

A more hands-on method of seeing that the horns aren't the whole simplicial set is to observe that $\Lambda^n_k\subset \Delta^n$ is generated by $n-1$-simplices, in the sense that all of its $m$-simplices for $m\geq n$ are degenerate. This is clear from the definition, and it seems hopefully clear that this condition is invariant under isomorphism (simplicial maps preserve degenerate simplices.) But of course $\Delta^n$ has a rather notable nondegenerate $n$-simplex, the identity map! Note that this isn't degenerate, since it doesn't factor through $n-1$.