The largest standard entry lies in a removable box

The entry $n$ is the largest entry of $t$, because $t$ is a bijection onto $\{1,\dots ,n\}$. If $(i,j+1)\in [\lambda ]$, then $t(i,j+1)>t(i,j)=n$ by [L1], which is impossible; hence $j={\lambda }_i$.

If $(i+1,j)\in [\lambda ]$, then $t(i+1,j)>t(i,j)=n$ by [L1], again impossible; hence $i=k$, or $i<k$ and ${\lambda }_i>{\lambda }_{i+1}$.

By steps 1.1 and 1.2 the node $b$ has the form $(i,{\lambda }_i)$ and satisfies ${\lambda }_i>{\lambda }_{i+1}$ with the convention ${\lambda }_{k+1}=0$, so $b$ is removable by [L2].

Let $\mu ⊢n-1$ be the partition with $[\mu ]=[\lambda ]\setminus \{b\}$, which exists by [L2], and let $t^{\mathrm{\prime }}$ be the restriction of $t$ to $[\mu ]$. Then $t^{\mathrm{\prime }}$ is a bijection $[\mu ]\to \{1,\dots ,n-1\}$, because $t$ is a bijection and the only node removed is the one carrying $n$.

Two nodes of $[\mu ]$ that are adjacent in a row or column of $[\mu ]$ are adjacent in $[\lambda ]$ and so satisfy the corresponding strict inequality in $t$; as their entries are unchanged by the restriction, the same strict inequality holds in $t^{\mathrm{\prime }}$. Hence $t^{\mathrm{\prime }}$ is a standard tableau of shape $\mu$, that is, a standard tableau of size $n-1$, and deleting the box $b$ occupied by $n$ has produced it. ∎