We will prove the theorem for fixed $k$ by induction on $n$. === Basis for the Induction === The case $n = 1$ ("Infinite Pigeonhole Principle"): In this case, $X^{\left({n}\right)}$ is the set of singletons of elements of $X$. So to ease exposition, we will identify $X^{\left({n}\right)}$ with $X$ and just speak of partitions of $X$. Let $P$ be a partition of $X$ into $k$ many components $X_1, \dots, X_k$. We have: :$X = X_1 \cup \cdots \cup X_k$ If each $X_i$ were finite, then $X$ would be a finite union of finite sets, and hence finite. Thus, at least one of the $X_i$ must be infinite. But taking $Y$ to be an infinite $X_i$ gives us the required subset for the theorem. This is the basis for the induction. === Induction Hypothesis === The induction hypothesis is that: For every partition $P$ of $X^{\left({n - 1}\right)}$ into $k$ many components :there is an infinite subset $Y \subseteq X$ such that: :each member of $Y^{\left({n - 1}\right)}$ is in the same component of $P$. It is to be proved that: For every partition $P$ of $X^{\left({n}\right)}$ into $k$ many components :there is an infinite subset $Y \subseteq X$ such that: :each member of $Y^{\left({n}\right)}$ is in the same component of $P$. === Induction Step === This is the induction step: Suppose the theorem holds for all $i < n$. It is to be proved that it holds for $n$. Since $X$ is infinite, there is an injection of $\N$ into $X$. All that is needed to be done to prove the theorem is to find an infinite subset of $X$ So, to ease exposition, $\N$ will be viewed as a subset of $X$. The proof will proceed by finding an infinite subset of $\N$. Write the components of $P$ as $S_1, \ldots, S_k$. For each $a \in \N$, let $P_a$ be the partition on $\left({\N \setminus \left\{ {a}\right\} }\right)^{ \left({n - 1}\right)}$ into $k$ components: :$S_{a, 1}, \dots, S_{a, k}$ defined by assigning: :$\left\{ {b_1, \ldots, b_{n - 1} }\right\} \in \left({\N \setminus \left\{ {a}\right\} }\right)^{\left({n - 1}\right)}$ to $S_{a, j}$ {{iff}}: : $\left\{ {a, b_1, \ldots, b_{n - 1} }\right\} \in \N^{\left({n}\right)}$ is in $S_j$ Note that the theorem holds for $P_a$ by the induction hypothesis. Now, we inductively define $a_0 < a_1 < \dots$ and infinite sets $X_0 \supseteq X_1 \supseteq \cdots$ such that $a_i$ is the smallest element of $X_i$. Let $a_0 = 0$, and let $X_0 = \N$. Once $X_i$ and $a_i$ are defined: Let $X_{i + 1}$ be an infinite subset of $X_i \setminus \left\{{a_i}\right\}$ satisfying the theorem for the partition $P_{a_i}$ of $n-1$ size subsets. This exists by the induction hypothesis. Let $a_{i + 1}$ be the smallest element of $X_{i+1}$. This exists by the Well-Ordering Principle. Note that $a_{i + 1} > a_i$ since $X_{i + 1}$ is a subset of $X_i$ not containing the smallest element $a_i$ of $X_i$. We constructed each $X_{i + 1}$ so that each element of $X_{i + 1}^{\left({n - 1}\right)}$ is in the same component of $P_{a_i}$. So, we may consider the sequence $\left({k_i}\right)_{i \in \N}$ where each $k_i$ is such that all elements of $X_{i + 1}^{\left({n - 1}\right)}$ are in the component $S_{a_i, k_i}$ of $P_{a_i}$. Since each $k_i$ is one of $1, \ldots, k$, but there are infinitely many $i \in \N$, this forms a partition of $\N$ to which we can apply the base case of this proof. This means that there is some $c$ among $1, \ldots, k$, there is an infinite set $I = \left\{ {i \in \N : k_i = c}\right\}$. Define $Y = \left\{ {a_i: i \in I}\right\}$. $Y$ is infinite since $I$ is infinite and each $a_i$ is distinct by construction. We now verify that each element of $Y^{\left({n}\right)}$ is in the same component $S_c$ of $P$. Let $\left\{ {y_1, \ldots, y_n}\right\} \in Y^{\left({n}\right)}$. We may take $y_1 < \cdots < y_n$ by relabeling if necessary. There is some $i \in I$ for which $y_1 = a_i$. By construction: :$y_1 = a_i$ is the least element of $X_i$ :each $y_2, \dots, y_n$ is in $X_i$ and: :each element of $X_i^{\left({n - 1}\right)}$ is in the component $S_{a_i, c}$ of the partition $P_{a_i}$. But then, by definition of $P_{a_i}$, since $\left\{ {y_2, \ldots, y_n}\right\} \in S_{a_i , c}$, we have: :$\left\{{y_1, y_2, \ldots, y_n}\right\} = \left\{ {a_i, y_2, \ldots, y_n}\right\} \in S_c$ {{qed}}