Euler proof of the formula involving factorial?

Question

Let me be formal and write the formula

Euler's Formula: Let $a$ and $n$ be nonnegative integers with $a\geq n.$ Then $$n!=a^n-\binom{n}{1}(a-1)^n+\binom{n}{2}(a-2)^n-\binom{n}{3}(a-3)^n+ > \cdot\cdot\cdot+(-1)^n\binom{n}{n}(a-n)^n$$

I've got this formula from the thesis "SELECTED PROOFS OF FERMAT'S LITTLE THEOREM AND WILSON'S THEOREM" by CAROLINE LAROCHE TURNAGE

The author scribbled a combinatorial proof. She also wrote that Euler originally proved this formula by induction. I was curious about that and I've tried a lot to prove the formula by induction, however I couldn't succeed. Here is my question: Is there anyone who could show me Euler's proof of this formula? Also, I'd be very glad if you provided me with a reliable resource other than mine.

http://math.stackexchange.com/questions/591350/factorial-as-difference-of-powers-sum-r-0n-binomnr-1rl-rn-n — lab bhattacharjee, Jun 23 '14 at 08:26
lab bhattacharjee there is no inductive proof with the link you gave me. — user159640, Jun 23 '14 at 08:30
This is true also for $a<n$ (it's a polynomial in $a$). Actually, it is the $n^{\text{th}}$ discrete derivative of the polynomial $X^n$. — Tom-Tom, Sep 15 '14 at 14:13

Tom-Tom · Answer 1 · 2014-09-17T07:45:54.943

Define $\Delta$ the discrete differential operator by its action on $f:\mathbf Z\to\mathbf R$: $$\Delta f(x)=f(x)-f(x-1).$$ It is clear that for every polynomial $P\in\mathbf R[X]$ the degree of $\Delta P(X)$ is the same as the degree of $P'$. Moreover, since $X^n-(X-1)^n=nX^{n-1}+Q_n(X)$, where $Q_n$ is a polynomial of degree at most $n-2$, the leading coefficient of $\Delta P$, that I note $a(\Delta P)$, is equal to $(\deg P)a(P)$.

Let us now compute $\Delta^n$. We will show the following expression by induction. $$\Delta^nP(X)=\sum_{k=0}^n(-1)^k\binom nkP(X-k).\tag 1$$

For $n=1$ it is the definition of $\Delta$. Suppose we have established (1) for some $n$. We will use the fact that $\Delta^{n+1}P=\Delta^n(\Delta P)$ (it's a little tricky, since intuitively one would use $\Delta^{n+1}P=\Delta\left(\Delta^nP\right)$.) $$ \Delta^{n+1}P=\sum_{k=0}^n(-1)^k\binom nk\left[P(X-k)-P(X-k-1)\right].$$ Now group the coefficient in front of $P(X-k)$ $$\Delta^{n+1}P=\sum_{k=0}^n(-1)^k\binom nk P(X-k)-\sum_{k=1}^{n+1}(-1)^{k-1}\binom n{k-1} P(X-k)$$ and we use the addition of binomials $\binom n{k-1}+\binom nk=\binom{n+1}k$ to conclude.

Finally, to answer your question we note that $\Delta^n(X^n)$ is a polynomial of degree equal to zero and of leading coefficient $a(\Delta^n(X^n))=n!$. Therefore $\Delta^n(X^n)=\Delta^n(X^n)(a)=n!$.

Euler proof of the formula involving factorial?

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