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,

(1)

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 $$f(x) \equiv f(a)+f'(a)(x\!-\!a)+f''(a)\frac{(x\!-\!a)^2}{2!}+\ldots+ f^{(\mu)}(a)\frac{(x\!-\!a)^{\mu}}{\mu!};$$ 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

(2)

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

(3)

(4)

We choose

(5)

For determining $F(0)$ we need, according to lemma 2, to expand $f(x)$ by the powers of $x$ , getting $$f(x) = \frac{1}{(p\!-\!1)!}[(-1)^{np}n!^px^{p-1}+A_1x^p+A_2x^p+1+\ldots]$$ 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 $$F(0) = \frac{1}{(p\!-\!1)!}[(-1)^{np}n!^p(p\!-\!1)!+A_1p! +A_2(p\!+\!1)!+\ldots] = (-1)^{np}n!^p+pK_0,$$ 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 $$f(x) = \frac{1}{(p\!-\!1)!}[B_p(x\!-\!\nu)^p+B_{p+1}(x\!-\!\nu)^{p+1}+\ldots],$$ where the $B_i$ 's are integers. Using the lemma 2 then gives the result $$F(\nu) = \frac{1}{(p\!-\!1)!}[p!B_p+(p\!+\!1)!B_{p+1}+\ldots] = pK_{\nu},$$ 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 $c_0F(0)+pK$ 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 $c_0F(0)$ . 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 non-zero 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 $$|f(x)| < \frac{1}{(p\!-\!1)!}n^{p-1}(n^n)^p = n^n\cdot\frac{(n^{n+1})^{p-1}}{(p\!-\!1)!}.$$ Because $\xi_1$ , ..., $\xi_n$ all are between 0 and $n$ (cf. (3)), we especially have $$|f(\xi_{\nu})| < n^n\cdot\frac{(n^{n+1})^{p-1}}{(p\!-\!1)!} \quad\forall \,\nu = 1,\,2,\,\ldots,\,n.$$ 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 $$(1\!+\!2\!+\!\ldots\!+\!n)ce^nn^n\cdot\frac{(n^{n+1})^{p-1}}{(p\!-\!1)!} = \frac{n(n\!+\!1)}{2}c(en)^n\cdot\frac{(n^{n+1})^{p-1}}{(p\!-\!1)!}.$$ 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.

Bibliography

1

ERNST LINDELÖF: Differentiali- ja integralilasku ja sen sovellutukset I. WSOY, Helsinki (1950).