Prove that at least one of $k$ consecutive integers is divisible by $k$.
We first consider the case $k=1$. In this case all inetegers are divisible by 1, an so the statement is trivially true.
We now consider the remaining $k \ge 2$.
A sequence of $k$ consecutive integers, beginning at $n$ is
$$ n,\; n + 1, \; n + 2, \; \ldots , \; n + (k - 1) $$
If we write $n = kq + r$ using the Division Algorithm, where $0 \le r < k$, we have
$$ kq + r,\; kq + r + 1, \; kq + r + 2, \; \ldots , \; kq + r + (k - 1) $$
Since $kq \equiv 0 \pmod k$, this sequence modulo $k$ is
$$ \boxed { r, \; r + 1, \; r + 2, \; \ldots , \; r + (k - 1) \pmod k } $$
For a given $r$ in $0 \le r \le k-1$, each element in the sequence is
$$ r + j $$
where $0 \le j \le k-1$.
We'll split $r$ into two cases, $r=0$ and $r \ge 1$.
In the case $r=0$, the first element of the sequence is then $j=0$, that is $0 \pmod k$, which is divisible by $k$.
In the case $r \ge 1$, we ask whether the sequence contains an element such that $r + j = k \equiv 0 \pmod k$. We can see that for any $r$ in the range $1 \le r \le k-1$ there is always a value of $j$ in the range $0 \le j \le k-1$ so that $r + j = k$. See below for a justification.
Because $ k \equiv 0 \pmod k$, each sequence has an element that is divisible by $k$.
So in both cases, there is an element in the sequence that is divisible by $k$.
We have shown that at least one of $k$ consecutive integers is divisible by $k$.
We want to show that for any $r$ in the range $1 \le r \le k-1$ there is always a value of $j$ in the range $0 \le j \le k-1$ such that $r + j = k$. Let's do this by induction on $k$.
Let $P(k)$ be the statement
For any $r$ in the range $1 \le r \le k-1$ there is always a value of $j$ in the range $0 \le j \le k-1$ such that $r + j = k$.
We need to show both the base case $P(k)$ and the inductive step $P(k) \implies P(k+1)$.
The base case is $P(2)$:
For any $r$ in the range $1 \le r \le 1$ there is always a value of $j$ in the range $0 \le j \le 1$ such that $r + j = 2$.
Here $r$ can only be 1, and $j$ can be 0 or 1. If we choose $j=1$, we have $r+k = 1 + 1 = 2$. So the base case is true.
The induction step is $P(k) \implies P(k+1)$. We assume the induction hypothesis $P(k)$, and show $P(k+1)$:
For any $r$ in the range $1 \le r \le k$ there is always a value of $j$ in the range $0 \le j \le k$ such that $r + j = k+1$.
The induction hypothesis tells us that there is an $r$ in the range $[1,k-1]$ and a $j$ in the range $[0,k-1]$ such that $r + j = k$. If we can pick $r$ from the range extended by one $[1,k]$ then we can have $r + j = k+1$. Alternatively, if we pick $j$ from the range extended by one $[0,k]$ then we can have $r + j = k+1$. And so $P(k+1)$ is true.
By induction we have shown that for any $r$ in the range $1 \le r \le k-1$ there is always a value of $j$ in the range $0 \le j \le k-1$ such that $r + j = k$.