Applying Induction on Higher Order Directional Derivatives Formula 
If $f$ is a function of two variables having continuous partials of order $k$ on some open set $S\subset \mathbb{R}^{2}$, show that $f^{(k)}(\mathbf{x};\mathbf{t})=\sum_{r=0}^{k}\binom{k}{r}t_{1}^{r}t_{2}^{k-r}D_{p_{1},p_{2},...,p_{k}}f(\mathbf{x})$ if $\mathbf{x}\in S$, $\mathbf{t}=(t_{1},t_{2})$, where in the rth term, we have $p_{1}=\cdots =p_{r}=1$ and $p_{r+1}=\cdots =p_{k}=2$. (Apostal)

I wanted to use induction with base case $k=2$ (I easily showed this), and the inductive hypothesis is assumed for $k=n\geq 2$. So far in the inductive step, I got the following:

$f^{(n+1)}(\mathbf{x};\mathbf{t})=t_{1}D_{1}f^{(n)}(\mathbf{x};\mathbf{t})+t_{2}D_{2}f^{(n)}(\mathbf{x};\mathbf{t})$

After applying the inductive hypothesis, I am not sure how to combine such a sum. I am guessing I have to change the summation bounds and use combinatorical identities to get the result I want. But I am not very good with this, so any advice or tips is appeciated!
 A: I'm gonna change notation from the unwieldy $D_{p_1\dots p_k}$ to $\partial_1^r \partial_2^{k-r}$, which is easier to manipulate algebraically. I'll also drop the $(\mathbf x;\mathbf t)$ to save space because its obvious where $f$ or $f^{(k)}$ are evaluated at.
The only combinatorial identity you need is the pascal triangle formula, 
$$\left(\begin{array}{l}{n} \\ {k}\end{array}\right)=\left(\begin{array}{c}{n-1} \\ {k}\end{array}\right)+\left(\begin{array}{l}{n-1} \\ {k-1}\end{array}\right)$$
The idea is very similar to the proof of the binomial theorem, which I suggest you study.
\begin{align}
f^{(n+1)}(\mathbf{x};\mathbf{t})
&=t_{1}\partial_{1}f^{(n)}(\mathbf{x};\mathbf{t})+t_{2}\partial_{2}f^{(n)}(\mathbf{x};\mathbf{t})
\\
&=t_{1}\partial_1\sum_{r=0}^{n}\binom{n}{r}t_1^r t_2^{n-r}\partial_1^r \partial_2^{n-r}f
+t_2\partial_2\sum_{r=0}^{n}\binom{n}{r}t_1^r t_2^{n-r}\partial_1^r \partial_2^{n-r}f
\\
&=\sum_{r=0}^{n}\binom{n}{r}t_1^{r+1} t_2^{n-r}\partial_1^{r+1} \partial_2^{n-r}f
 +\sum_{r=0}^{n}\binom{n}{r}t_1^r t_2^{n+1-r}\partial_1^r \partial_2^{n+1-r}f
\\
&=\sum_{r=0}^{n}\binom{n}{r}t_1^{r+1} t_2^{n+1-(r+1)}\partial_1^{r+1} \partial_2^{(n+1)-(r+1)}f
 +\sum_{r=0}^{n}\binom{n}{r}t_1^r t_2^{n+1-r}\partial_1^r \partial_2^{n+1-r}f
\\
&=\sum_{R=1}^{n+1}\binom{n}{R-1}t_1^{R} t_2^{n+1-R}\partial_1^{R} \partial_2^{(n+1)-R}f
 +\sum_{r=0}^{n}\binom{n}{r}t_1^r t_2^{n+1-r}\partial_1^r \partial_2^{n+1-r}f
\\
&=\sum_{r=1}^{n}\binom{n}{r-1}t_1^{r} t_2^{n+1-r}\partial_1^{r} \partial_2^{(n+1)-r}f
 +\sum_{r=1}^{n}\binom{n}{r}t_1^r t_2^{n+1-r}\partial_1^r \partial_2^{n+1-r}f
\\
& \quad + \binom{n}{n}t_1^{n+1}\partial_1^{n+1}f +\binom{n}{0} t_2^{n+1}\partial_2^{n+1}f \\
&= \sum_{r=1}^{n}\left[\binom{n}{r}+\binom{n}{r-1}\right]t_1^r t_2^{n+1-r}\partial_1^r \partial_2^{n+1-r}f  + \binom{n}{n}t_1^{n+1}\partial_1^{n+1}f + \binom{n}{0} t_2^{n+1}\partial_2^{n+1}f 
\\
&= \sum_{r=1}^{n}\binom{n+1}{r}t_1^r t_2^{n+1-r}\partial_1^r \partial_2^{n+1-r}f  + \binom{n+1}{n+1}t_1^{n+1}\partial_1^{n+1}f + \binom{n+1}{0} t_2^{n+1}\partial_2^{n+1}f 
\\
&= \sum_{r=0}^{n+1}\binom{n+1}{r}t_1^r t_2^{n+1-r}\partial_1^r \partial_2^{n+1-r}f
\end{align}
