Background (from digital signal processing):
The $\delta$ function is defined as below:
$$\delta[n]=\begin{cases} 1 \quad \mbox{if} \hspace{.4em} n=0\\ 0 \quad \mbox{o.w.} \end{cases}$$
In an LTI (Linear Time-Invariant) system, we denote the response to the $\delta$ function by $h$.
Question:
Assume that $h$ is defined as below:
$$h(n)=\sum_{k=1}^p a_k h[n-k]+G \delta[n]$$
where $a_k$'s are some defined coefficients, and $G$ is another fixed coefficient like them. $p$ also a fixed value.
Also, let the auto-correlation function $\hat{R}(m)$ be defined as follows:
$$\hat{R}(m)=\sum_{n=0}^{\infty}h[n]h[n+m]$$
Prove that $\hat{R}(m)=\hat{R}(-m)$
My try:
I managed to define a new variable $v=n+m$. Since $n$ was from $0$ to $\infty$, $v$ will be from m to $\infty$. Re-writing $\hat{R}(m)$ using the variable, we have:
\begin{align} \hat{R}(m)&=\sum_{v=m}^{\infty}h[v-m]h[v]\\ &=\sum_{v=0}^{\infty}h[v-m]h[v]-\sum_{v=0}^{m-1}h[v-m]h[v]\\ &=\sum_{n=0}^{\infty} h[n-m]h[n]-\sum_{n=0}^{m-1}h[n-m]h[n]\\ &=\hat{R}(-m)-\sum_{n=0}^{m-1}h[n-m]h[n] \end{align}
Now I need to somehow prove that $\sum_{n=0}^{m-1}h[n-m]h[n]$ is equal to $0$. However, I do not know how. It looks like a deadend.
Note: Denoting the functions like $f[n]$ instead of $f(n)$ in this context is due to the fact that we are working with digital signals where we assume that the domain consists only of the integers.