7. Given positive integers $n, p$ greater than 1, for any positive integer, if its $p$-ary representation contains no digit zero, then it is called “$p$-full”. Then, the number of $p$-full natural numbers among $1,2, \cdots, p^{n}$ is $\qquad$
7. $\left\{\begin{array}{ll}n, & p=2 ; \\ \frac{(p-1)^{n+1}-p+1}{p-2}, & p>2 .\end{array}\right.$

Obviously, the number of $p$-full natural numbers with $k(1 \leqslant k \leqslant n)$ digits in $p$-ary representation is $(p-1)^{k}$. Therefore, the number of $p$-full natural numbers we are looking for is
$$
\sum_{k=1}^{n}(p-1)^{k}=\left\{\begin{array}{ll}
n, & p=2 ; \\
\frac{(p-1)^{n+1}-p+1}{p-2}, & p>2
\end{array}\right.
$$