This theorem was first proved by Hermite in 1873.  The below proof is near the one given by Hurwitz.  We at first derive a couple of auxiliary results.

Let $f(x)$ be any polynomial of degree $\mu$ and $F(x)$ the sum of its derivatives,

consider the product  $\Phi(x):=e^{{-x}}F(x)$.  The derivative of this is simply

Applying the mean value theorem to the function $\Phi$ on the interval with end points 0 and $x$ gives

which implies that  $F(0)=e^{{-x}}F(x)+e^{{-\xi}}f(\xi)x$.  Thus we obtain the

Lemma 1.  $F(0)e^{x}=F(x)+xe^{{x-\xi}}f(\xi)$ ($\xi$ is between 0 and $x$)

When the polynomial $f(x)$ is expanded by the powers of $x\!-\!a$, one gets

comparing this with (1) one gets the

Lemma 2.  The value $F(a)$ is obtained so that the polynomial $f(x)$ is expanded by the powers of $x\!-\!a$ and in this expansion the powers $x\!-\!a$, $(x\!-\!a)^{2}$, …, $(x\!-\!a)^{{\mu}}$ are replaced respectively by the numbers 1!, 2!, …, $\mu!$.

Now we begin the proof of the theorem.  We have to show that there cannot be any equation

with integer coefficients $c_{i}$ and at least one of them distinct from zero.  The proof is indirect.  Let’s assume the contrary.  We can presume that  $c_{0}\neq 0$.

For any positive integer $\nu$, lemma 1 gives

By virtue of this, one may write (2), multiplied by $F(0)$, as

We shall show that the polynomial $f(x)$ can be chosen such that the left side of (4) is a non-zero integer whereas the right side has absolute value less than 1.

We choose

where $p$ is a positive prime number on which we later shall set certain conditions.  We must determine the corresponding values $F(0)$, $F(1)$, …, $F(n)$.

For determining $F(0)$ we need, according to lemma 2, to expand $f(x)$ by the powers of $x$, getting

where $A_{1},\,A_{2},\,\ldots$ are integers, and to replace the powers $x^{{p-1}}$, $x^{p}$, $x^{{p+1}}$, … with the numbers $(p\!-\!1)!$, $p!$, $(p\!+\!1)!$, …  We then get the expression

in which $K_{0}$ is an integer.

We now set for the prime $p$ the condition  $p>n$.  Then, $n!$ is not divisible by $p$, neither is the former addend $(-1)^{{np}}n!^{p}$.  On the other hand, the latter addend $pK_{0}$ is divisible by $p$.  Therefore:
($\alpha$) $F(0)$ is a non-zero integer not divisible by $p$.

For determining $F(1)$, $F(2)$, …, $F(n)$ we expand the polynomial $f(x)$ by the powers of $x\!-\!\nu$, putting  $x:=\nu\!+\!(x\!-\!\nu)$.  Because $f(x)$ contains the factor $(x\!-\!\nu)^{p}$, we obtain an expansion of the form

where the $B_{i}$’s are integers.  Using the lemma 2 then gives the result

with $K_{{\nu}}$ a certain integer.  Thus:
($\beta$) $F(1)$, $F(2)$, …, $F(n)$ are integers all divisible by $p$.

So, the left hand side of (4) is an integer having the form $c_{0}F(0)+pK$ with $K$ an integer.  The factor $F(0)$ of the first addend is by ($\alpha$) indivisible by $p$.  If we set for the prime $p$ a new requirement  $p>|c_{0}|$,  then also the factor $c_{0}$ is indivisible by $p$, and thus likewise the whole addend $c_{0}F(0)$.  We conclude that the sum is not divisible by $p$ and therefore:
($\gamma$) If $p$ in (5) is a prime number greater than $n$ and $|c_{0}|$, then the left side of (4) is a nonzero integer.

We then examine the right hand side of (4).  Because the numbers $\xi_{1}$, …, $\xi_{n}$ all are positive (cf. (3)), so the exponential factors $e^{{1-\xi_{1}}}$, …, $e^{{n-\xi_{n}}}$ all are $<e^{n}$.  If  $0<x<n$, then in the polynomial (5) the factors $x$, $x\!-\!1$, …, $x\!-\!n$ all have the absolute value less than $n$ and thus

Because $\xi_{1}$, …, $\xi_{n}$ all are between 0 and $n$ (cf. (3)), we especially have

If we denote by $c$ the greatest of the numbers $|c_{0}|$, $|c_{1}|$, …, $|c_{n}|$, then the right hand side of (4) has the absolute value less than

But the limit of $\frac{(n^{{n+1}})^{{p-1}}}{(p\!-\!1)!}$ is 0 as  $p\to\infty$, and therefore the above expression is less than 1 as soon as $p$ exeeds some number $p_{0}$.

If we determine the polynomial $f(x)$ from the equation (5) such that the prime $p$ is greater than the greatest of the numbers $n$, $|c_{0}|$ and $p_{0}$ (which is possible since there are infinitely many prime numbers), then the left side of (4) is a non-zero integer and thus $\geqq 1$, whereas the right side having the absolute value $<1$.  The contradiction proves that the theorem is right.