Let $\lambda$ be in $\mathbb F_{p^\ell}\setminus \{0\}$ where $p$ is prime and $\ell \in \mathbb N_{\ge 1}$, and $j,k$ be three integers.
In order to calculate the multiplicative order of $X+\lambda$ in $(\mathbb F_{p^\ell}/(X^t))^*$ (which I know is equal to $p^\delta\mathrm{ord}(\lambda)$ where $\delta$ is such that $p^{\delta-1}<t\leqslant p^\delta$), I need to prove that there exists $S\in \mathbb F_{p^\ell}[X]$ such that $S(0)=0$, such that
$$(X+\lambda)^{jp^k}=\lambda^{jp^k}+j\lambda^{(j-1)p^k}X^{p^k}(1+S(X)).$$
I try to use Newton's binomial theorem, but then I am stuck with proving that
$$\binom {jp^k}{p^k}=j$$
and
$$\binom {jp^k}{x}=0\in \mathbb F_{p^\ell}$$
if $p^k$ does not divide $x$.
Am I going the wrong way ? How should I prove this result ?