Let us interpret this process as operations on a graph. So, we have an initial graph and are allowed to sequentially delete vertices that are connected to a vertex of degree $1$. We must prove that acting like that it is impossible to be left with isolated vertices with different cardinalities. Consider the sequence of operations backwards. We start from an initial graph $G_0$ with $m$ isolated vertices. At $i$-th step, $i=1,2,\dots$ we are allowed to add to $G_{i-1}$ a new vertex $v$. We connect $v$ to some $v'\in V(G_0)$ that is still isolated in $G_{i-1}$. We also may connect $v$ to some subset (which may be empty) of vertices in $G_{i-1}$ thus obtaining a graph $G_i$. Let us refer to this operation as "adding a vertex $v$". We will prove that it's impossible to start with two sets of isolated vertices $G_0$ and $G'_0\,,\, |G_0|\neq |G'_0|$ and after a sequence of operations to obtain the same graph $G.$ That is, knowing that $G$ is obtained by a sequence of operations $$G_0\to G_1\to\dots\to G_k=G \qquad (1)$$we can recover $|G_0|$ uniquely. We will prove it by induction on the number of vertices in $G$ (on the complexity of $G$). It's trivial when $G$ consists of $1,2$ or $3$ vertices. Consider a graph $G$ with $n$ vertices for which a sequence like in $(1)$ does exist. Assume that we can recover the last operation in the sequence $(1)$, i.e. there exists only one vertex $v$ in $G$ that is connected to a vertex of degree $1$. In this case we remove $v$ and apply the induction hypothesis. Assume now there exist at least two vertices $v_1,v_2$ such that $v_1$ is connected to $v'_1,$ $v_2$ is connected to $v'_2$ and $d(v'_1)=d(v'_2)=1$. Apparently $v'_1\neq v'_2$. Here is a crucial observation. Suppose a vertex $v$ is connected to a vertex $v'$ in $G$ and $d(v')=1.$ It means, at some stage in $(1)$ there is an operation "adding a vertex $v$". Then we can "move" this operation at the final step and the result, i.e. the graph $G$, will be the same. So, we can assume that "adding a vertex $v_1$" and "adding a vertex $v_2$" are the last two operations. If we remove $v_1$ we get a graph with less vertices and by the induction hypothesis we can recover the number of vertices of its initial graph. The same is true if we remove $v_2$ first. We claim that the two obtained numbers are equal. Indeed, if they are not, and we remove both vertices $v_1,v_2$, we get a graph $G'$ and the cardinality of its initial state $G'_0$ would be equal to two different numbers, contradiction.