Verify taking the derivative of this polynomial

I've been splitting my head over this, and most likely it's obvious but I can't see it. Taken from Understanding Analysis, 2nd Edition, Stephen Abbott.

Exercise $$\mathbf{6.6.9}$$ (Cauchy's Remainder Theorem) Let $$f$$ be differentiable $$N+1$$ times on $$(-R,R)$$. For each $$a\in(-R,R)$$, let $$S_N(x,a)$$ be the partial sum of the Taylor series for $$f$$ centered at $$a;$$ in other words, define $$S_N(x,a)=\sum_{n=0}^{N}{c_n(x-a)^n} \text{ where } c_n=\frac{f^{(n)}(a)}{n!}.$$ Let $$E_N(x,a)=f(x)-S_N(x,a).$$ Now fix $$x\neq0$$ in $$(-R,R)$$ and consider $$E_N(x,a)$$ as a function of $$a$$.

Explain why $$E_N(x,a)$$ is differentiable with respect to $$a$$, and show $$E'_N(x,a)=-\frac{f^{(N+1)}(a)}{N!}(x-a)^N.$$

My attempt. $$E_N$$ is differentiable w.r.t $$a$$ because $$f(x)$$ is constant, and $$S_N$$ is a polynomial in $$a$$ and hence differentiable. In fact we can say $$E_N$$ is infinitely differentiable. Now, \begin{align} E'_N=-S'_N &=-\sum_{n=1}^{N}{nc_n(x-a)^{n-1}(-1)}\\ &=\sum_{n=1}^{N}{nc_n(x-a)^{n-1}}\\ &=\sum_{n=1}^{N}{n\frac{f^{(n)}(a)}{n!}(x-a)^{n-1}}\\ &=\sum_{n=1}^{N}{\frac{f^{(n)}(a)}{(n-1)!}(x-a)^{n-1}}\\ \end{align} I'm stuck here. How to proceed?

When you differentiate $$S_N$$ term-wise, you can't ignore that $$c_n$$ depends on $$a$$ (and so $$S_N(x, a)$$ isn't polynomial in $$a$$).
Correct way will be $$E_N' = \frac{\partial}{\partial a} \left(- \sum_{n = 0}^N\frac{f^{(n)}(a)}{n!}(x - a)^n\right) \\ = - f^{(1)}(a) - \sum_{n=1}^N \frac{\partial}{\partial a} \left(\frac{f^{(n)}(a)}{n!}(x - a)^n\right)\\ = -f^{(1)}(a) - \sum_{n=1}^N \frac{f^{(n + 1)}(a)}{n!}(x - a)^n - \sum_{n=1}^N-\frac{f^{(n)}(a)}{(n - 1)!}(x - a)^{n - 1} \\ = -f^{(1)}(a) - \sum_{n=1}^N \frac{f^{(n + 1)}(a)}{n!}(x - a)^n + \sum_{n = 0}^{N - 1}\frac{f^{(n + 1)}(a)}{n!}(x - a)^n\\ = \color{red}{-f^{(1)}(a)} - \color{blue}{\sum_{n=1}^{N - 1} \frac{f^{(n + 1)}(a)}{n!}(x - a)^n} - \frac{f^{(N + 1)}(a)}{N!}(x - a)^N + \color{red}{f^{(1)}(a)} + \color{blue}{\sum_{n = 1}^{N - 1}\frac{f^{(n + 1)}(a)}{n!}(x - a)^n}\\ = -\frac{f^{(N + 1)}(a)}{N!}(x - a)^N$$