1.置换不变性:若$(i_0,i_1,\cdots,i_k)$是$0,1,2,\cdots,k$的任一置换，则有
\begin{equation}
f[t_{i_0},t_{i_1},\cdots,t_{i_k}]=f[t_0,t_{1},\cdots,t_{k}]\end{equation}证明:我们使用数学归纳法证明.当$k=1$时，命题显然成立.设当$k\leq p$时，命题都成立,则$k=p+1$时，
\begin{equation}\label{eq:2}
f[t_{i_0},t_{i_1},\cdots,t_{i_{p+1}}]=\frac{f[t_{i_0},t_{i_1},\cdots t_{i_p}]-f[t_{i_1},t_{i_2},\cdots,t_{i_{p+1}}]}{t_{i_0}-t_{i_{p+1}}}\end{equation}如果置换发生在$t_{i_0},t_{i_1},\cdots,t_{i_p}$之间，或者是$t_{i_1},t_{i_2},\cdots,t_{i_{p+1}}$之间，将不会对\ref{eq:2}的值产生影响(根据归纳假设).如果置换发生在$t_{i_0}$和$t_{i_{p+1}}$之间，易得也不会对\ref{eq:2}的值产生影响.综上，可得当$k=p+1$时，命题也成立(为什么?).根据数学归纳法，命题成立.
2.线性性: \begin{align*}
f[x_0,x_1,\cdots,x_k]=\sum_{j=0}^k \frac{f(x_j)}{\prod_{i\neq j;0\leq i\leq k}(x_j-x_i)} \end{align*}证明:我们不准备使用严格的数学归纳法证明,而只是一直做下去，产生一种归纳的感觉.当$k=1$时，
\begin{align*} f[x_0,x_1]=\frac{f(x_0)}{x_0-x_1}+\frac{f(x_1)}{x_1-x_0}\end{align*}当$k=2$时，\begin{align*} f[x_0,x_1,x_2]=\frac{f[x_0,x_1]-f[x_1,x_2]}{x_0-x_2}=\frac{\frac{f(x_0)-f(x_1)}{x_0-x_1}-\frac{f(x_1)-f(x_2)}{x_1-x_2}}{x_0-x_2}=\frac{f(x_0)}{(x_0-x_1)(x_0-x_2)}+\frac{f(x_1)}{(x_1-x_0)(x_1-x_2)}+\frac{f(x_2)}{(x_2-x_0)(x_2-x_1)}\end{align*}当$k=3$时，\begin{align*} f[x_0,x_1,x_2,x_3]&=\frac{f[x_0,x_1,x_2]-f[x_1,x_2,x_3]}{x_0-x_3}\\&=\frac{[\frac{f(x_0)}{(x_0-x_1)(x_0-x_2)}+\frac{f(x_1)}{(x_1-x_0)(x_1-x_2)}+\frac{f(x_2)}{(x_2-x_0)(x_2-x_1)}]-[\frac{f(x_1)}{(x_1-x_2)(x_1-x_3)}+\frac{f(x_2)}{(x_2-x_1)(x_2-x_3)}+\frac{f(x_3)}{(x_3-x_1)(x_3-x_2)}]}{x_0-x_3}\\&=\frac{f(x_0)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}+\frac{f(x_1)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}+\frac{f(x_2)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}+\frac{f(x_3)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}\end{align*}……就这样一直下去，易知命题成立.
3.若函数$f(t)$在$t_1$的邻域内具有直到$n$阶的连续导数，节点$t_1,t_2,\cdots,t_s$两两互异，则
\begin{align*}
f[t_1,\cdots,t_1\mbox{(n+1个 )},t_2,\cdots,t_s]=\frac{1}{n!}\frac{\partial^n}{\partial(t_1^n)}f[t_1,t_2,\cdots,t_s]\end{align*}
证明:这个命题过于普遍，我想仅仅通过不失一般性的特例来满足自己.不妨探索
\begin{align*}
f[t_1,t_2,t_3,t_4]\end{align*}在这条式子里，给$t_1$求一次导，会发生什么状况?也就是
\begin{align*}
\lim_{\varepsilon\neq 0;\varepsilon\to 0} \frac{f[t_1+\varepsilon,t_2,t_3,t_4]-f[t_1,t_2,t_3,t_4]}{(t_1+\varepsilon)-t_1}\end{align*}根据差商的定义，我们会发现，上式等于
\begin{align*}
f[t_1,t_1,t_2,t_3,t_4]\end{align*}现在给\begin{align*} f[t_1,t_2,t_3,t_4]\end{align*}中的$t_1$求两次导，即，给$f[t_1,t_1,t_2,t_3,t_4]$求一次导，
\begin{align*}
\lim_{\varepsilon\to 0;\varepsilon\neq 0}\frac{f[t_1+\varepsilon,t_1+\varepsilon,t_2,t_3,t_4]-f[t_1,t_1,t_2,t_3,t_4]}{(t_1+\varepsilon)-t_1}~~~~~\mbox{ 【1】}\end{align*}毫无疑问，我们会使用来处理这种事情.将上面的式子变为\begin{align*}
\lim_{\varepsilon\neq 0;\varepsilon\to 0} \frac{f[t_1+\varepsilon,t_1+\varepsilon,t_2,t_3,t_4]-f[t_1+\varepsilon,t_1,t_2,t_3,t_4]+f[t_1+\varepsilon,t_1,t_2,t_3,t_4]-f[t_1,t_1,t_2,t_3,t_4]}{t_1+\varepsilon-t_1}\end{align*}由于\begin{align*}
\frac{f[t_1+\varepsilon,t_1+\varepsilon,t_2,t_3,t_4]-f[t_1+\varepsilon,t_2,t_3,t_4,t_1]}{(t_1+\varepsilon)-t_1}=f[t_1+\varepsilon,t_1+\varepsilon,t_2,t_3,t_4,t_1]\end{align*}\begin{align*}
\frac{ f[t_1+\varepsilon,t_1,t_2,t_3,t_4]-f[t_1,t_2,t_3,t_4,t_1]}{(t_1+\varepsilon)-t_1}=f[t_1+\varepsilon,t_1,t_2,t_3,t_4,t_1]\end{align*}可见，【1】式可以变为\begin{align*} 2f[t_1,t_1,t_1,t_2,t_3,t_4]\end{align*} 如果我们给$f[t_1,t_2,t_3,t_4]$中的$t_1$求三次导,也就是给\begin{align*} 2f[t_1,t_1,t_1,t_2,t_3,t_4]\end{align*}求一次导.易得会变为\begin{align*} 6f[t_1,t_1,t_1,t_1,t_2,t_3,t_4]\end{align*}……可见，命题是有理由成立的.