Given a book with $100$ pages and a $100$ lemmas, prove that there is some lemma written on the same page as its index 
A book consists of 100 pages and contains 100 lemmas and some images. Each lemma is at most one page long and can't be split into two pages (it has to fit in one page). The lemmas are numbered from 1 to 100 and are written in ascending order. Prove that there must be at least one lemma written on a page with the same number as the lemma's number.

If lemma no. $1$ is written on page $1$, then it is proved. Let's assume lemma no. $1$ is written on page  $k, k>1$. Then on at least one page there must be $2$ lemmas. Let's assume that always on page $k+i$ we have lemma no. $i+1$ and so on. Then the last $100-k-i$ lemmas must fit on the last page, which means that there will be at least one lemma (number $100$) on page $100$.
But I don't know how to express it in a more mathematical way!
Any help?

Another proof, this time not by induction.
[Edit:  on being tidied up, this turned out to be basically an expanded version of the other proofs that don't use induction.]
Consider a book of $k$ lemmas and $k$ pages.
For $1\le n \le k$ and $ n\in\mathbb N$, let


*

*$p_n$ be the page number of the $n$th lemma  

*$D_n=n-p_n$ be the difference between the lemma number and its page number.


We are asked to prove that for some $n$, $D_n=0$.
Suppose we have assigned pages to the first $n$ lemmas. The next lemma can be on either the same page as lemma $n$ or a later page, i.e.
$$p_{n+1}\ge p_n$$
from which
$$D_{n+1}\le D_n+1$$
That is, $D_n$ can't increase by more than $1$ between one lemma and the next.
If the first lemma is on the first page ($p_1=1$), or the last lemma is on the last page ($p_k=k$), we already have a lemma on its own-numbered page. So we need only consider the case where $p_1>1$ and $p_k<k$, making $D_1<0$ and $D_k>0$.
$D_n$ must therefore increase from a negative value to a positive one somewhere between $n=1$ and $n=k$. But it can only increase in steps of $1$. So for $D_n$ to change sign, there must be a value of $n$ for which $D_n=0$, ie for which lemma $n$ is on page $n$.
There is therefore at least one lemma on a page with  the same number as the lemma.
