7. Given the polynomial
$$
\begin{array}{l}
(1+x)+(1+x)^{2}+(1+x)^{3}+\cdots+(1+x)^{n} \\
=b_{0}+b_{1} x+b_{2} x^{2}+\cdots+b_{n} x^{n},
\end{array}
$$

and it satisfies $b_{1}+b_{2}+\cdots+b_{n}=26$. Then a possible value of the positive integer $n$ is $\qquad$
Take $x=0$, we get $b_{0}=n$.
Take $x=1$, we get
$$
2+2^{2}+2^{3}+\cdots+2^{n}=b_{0}+b_{1}+b_{2}+\cdots+b_{n} \text {, }
$$

i.e., $\frac{2\left(1-2^{n}\right)}{1-2}=n+26$.
Thus, $2\left(2^{n}-1\right)=n+26$.
Therefore, $n=4$ is a possible value.