Every harmonic function is real analytic, Evans p.31-32 In L.Evans book 'Partial Differential Equations' theorem $10$ page $31$ proves that every harmonic function is real analytic. I am stuck in a certain point of the proof. He mentions:
To verify this, let us compute for each $N$ the remainder term  $$
\begin{align}
R_N(x) &= u(x)-  \sum_{k=0}^{N-1} { \sum_{|a|=k}{\dfrac{D^a u(x_0)}{a!}}(x-x_0)^a} \\&= \sum_{|a|=N} \dfrac{D^a u(x_0+t(x-x_0))}{a!} (x-x_0)^a   
\end{align}\tag{*}$$
for some $t \in [0,1]$, $t$ depending on $x$. We establish this formula by writing out the first $N$ terms and the error in the Taylor expansion about $0$ for the function of one variable
$$g(t):=u(x_0+t(x-x_0))\quad \text{at $t=1$.}$$
Question: Can someone provide a detailed proof of the second equality above in $(*)$?
My approach: As the hint suggests I computed using the chain rule that $$g^{(k)}(t)=\sum_{|a|=k} D^a u(x_0+t(x-x_0)) (x-x_0)^a $$ and so writing $$g(t)= \sum_{k=0}^{\infty} \frac{g^{(k)}(0)}{k!}t^k  $$ one gets by plugging in $t=1$ that $$u(x)= \sum_{k=0}^{\infty} { \sum_{|a|=k}{\dfrac{D^a u(x_0)}{k!}}(x-x_0)^a} $$
Then, I split the sum from $k=0$ to $N-1$ and from $N$ to $\infty$ but I don't know how to continue.
 A: There's a subtle error in your calculations. Calculating with the chain rule we get
$$g^{(k)} (t) = \sum_{j_1,...,j_k=1}^n u_{x_{j_1},...,x_{j_k}}(x_0+t(x-x_0))(x_{j_1}-x_{0j_1})\cdot...\cdot (x_{j_k}-x_{0j_k}).$$
From this we can't directly move into multi-index notation. Notice that due to smoothness the higher-order derivatives are independent of order, so certain terms in the sum get repeated. More precisely, the term
$u_{x_{i_1},...,x_{i_k}}(x_0+t(x-x_0))(x_{i_1}-x_{0i_1})\cdot...\cdot (x_{i_k}-x_{0i_k})$ gets repeated $\binom{k}{i_1...i_k}=\frac{k!}{i_1!\cdots i_k!}$ times. (This is the multinomial coefficient, and it's interpreted as the amount of ways to choose a sequence $(x_{j_1},...,x_{j_k})$ such that in that sequence $x_{i_1}$ appears $i_1$ times, etc.) From this we can write
$$
g^{(k)}(t)=\sum_{|\alpha|=k} \frac{k!}{\alpha!}D^{\alpha}u(x_0+t(x-x_0))(x-x_0)^\alpha,
$$
and thus the degree $N-1$ Taylor polynomial around $t=0$ is
$$
T_{N-1}(s)=
\sum_{k=0}^{N-1}\frac{1}{k!}\left(\sum_{|\alpha|=k}\frac{k!}{\alpha!}D^{\alpha}u(x_0)(x-x_0)^\alpha\right)s^k.\\
= \sum_{k=0}^{N-1}\sum_{|\alpha|=k}\frac{1}{\alpha!}D^\alpha u(x_0)(x-x_0)^\alpha
s^k.$$
Rest of the proof should be straightforward enough, just use the remainder formula at $s=1$.
