Question about Damgård–Jurik crypto system
The simple answer is that the element $(n+1)$ is of order $n^s$, i.e. $(n+1)^{n^s} \equiv 1 \bmod n^{s+1}$. Thus the first part can be written as $(1+n)^{jmd \bmod n^s}$. 

For the other part it is $(x^mr^{n^s})^d$, it is written in that way so that it is easier to see the setting that $d \equiv 0 \bmod \lambda$ ensures this part $(x^mr^{n^s})^d \equiv 1 \bmod n^{s+1}$. To see why, note that $x^\lambda \equiv 1 \bmod n^{s+1}$ and also $(r^{n^s})^\lambda \equiv 1 \bmod n^{s+1}$. The explaination is as the following:


We know that $x\in H$ where $H=\{g^{n^s} \bmod n^{s+1}|g\in Z_{n^{s+1}}^*\}$ is a subgroup of $Z_{n^{s+1}}^*$, we know $r \in Z_{n}^*$ and $Z_{n}^*$ is a subgroup of $Z_{n^{s+1}}^*$ (thus $r$ is also in $Z_{n^{s+1}}^*$).
We know the exponent (the least common multiple of the orders of all elements of the group) of $Z_{n^{s+1}}^*$ is $n^s\lambda$ (see here, $\lambda$ in the paper is a shorthand for $\lambda(n)$). In other words, $a^{n^s\lambda}\equiv 1 \bmod n^{s+1}$ holds for any element $a$ in $Z_{n^{s+1}}^*$ 
Therefore, $x^\lambda=g^{n^s\lambda}\equiv 1 \bmod n^{s+1}$, and $r^{n^s\lambda}\equiv 1 \bmod n^{s+1}$ 


It thus follows that if $d\equiv 0 \bmod \lambda$ (i.e. $d = k\lambda $ for some k), then $(x^mr^{n^s})^d \equiv 1 \bmod n^{s+1}$.