Problem: Let $X_1,X_2,\ldots \sim \text{Exp}(1)$. Denote $S_n=\sum_{k=1}^nX_k$, $N(x)=\inf\{ n>0\mid S_n>x\}$, $m=\mathbb{E}[N(x)]$. Prove by definition, $$\mathbb{E}[N(x)\mid X_1]= \begin{cases} m(x-X_1)+1 &,\text{ if } X_1\le x\\ 1 &,\text{ if } X_1>x \end{cases}=:g(x)$$
I'm taking a course in probability and we just defined conditional expectation. intuitive I understand why this is equal but I'm having trouble proving it by definition.
My approach:
Define $A=\{{X_1<s}\}$ an and from The π- theorem it is enough to show for all s, $\mathbb{E}[N(x)*1_A ]=\mathbb{E}[\mathbb{E}[N(x)|X_1]*1_A]$.
$s\le t:$
$\mathbb{E}[N(t)*1_A]=\mathbb{E}[N(t)*1_{X_1<s}]=\sum_{k=0}^\infty k\mathbb{P}(N(t)=k,\ X_1<s)$.
We'll notice that $$\mathbb{P}(N(t)=k,\ X_1<s)=\mathbb{P}(\forall m\le k:\ S_m<t,\ S_k\ge t,\ X_1<s)$$
$$=\mathbb{P}(\forall m\le k:\ X_2+...+X_m<t-X_1,\ X_2+...+X_k\ge t-X_1,\ X_1<s)$$
$$=\mathbb{P}(N(t-X_1)+1=k,X_1<s)$$
And so $$\mathbb{E}[N(t)*1_A]=\mathbb{E}[N(t)*1_{X_1<s}]=\sum_{k=0}^\infty k\mathbb{P}(N(t)=k,\ X_1<s)=\sum_{k=0}^\infty k\mathbb{P}(N(t-X_1)+1=k,X_1<s)=\mathbb{E}[(1+N(t-X_1))1_{X_1<s}]$$
$s> t$:
I tried instead of working with $\{X_1<s\}$ to work with $\{X_1\ge s\}$
$$\mathbb{E}[N(x)*1_{X_1<s} ]=\mathbb{E}[N(x)*(1-1_{X_1\ge s})]=\mathbb{E}[N(x)]-\mathbb{E}[N(x)*1_{X_1\ge s})]=\mathbb{E}[N(x)]-\mathbb{P}(X_1\ge s)$$
$$\mathbb{E}[g(x)1_{X_1< s}]=\mathbb{E}[g(x)]-\mathbb{P}(X_1\ge s)$$
But then I get that I need to prove $\mathbb{E}[N(x)]=\mathbb{E}[g(x)]$. Which is the original claim I want to prove.
How do I continoue from here?
Thank you