For $X_t=\sum_{j=1}^d \int_0^t f_j(s)dB_s^j - \frac{1}{2}\int_0^t |f(s)|^2ds$, $\tau_n=\inf\{s\ge 0: |\exp(X_s)|\ge n\}\wedge n$ goes to $\infty$.

Question

This is part of a proof of a theorem on Doléans-Dade exponentials by Rene Schilling's Brownian Motion.

Theorem Let $B_t$ be a $d$-dimensional Brownian Motion and $f=(f_1, \dots, f_d)$ be a $d$-dimensional progressively measurable process such that $f_j\in L^2_{\mathscr{P}}(\lambda_T \times \mathbb{P})$ for all $T>0$ and $j=1,\dots d$.

Let $X_t=\sum_{j=1}^d \int_0^t f_j(s)dB_s^j - \frac{1}{2}\int_0^t |f(s)|^2ds$.

Then, $M_t=\exp(X_t)$ is a martingale if, and only if, $\mathbb{E} M_t = 1$.

Question. Now let $\tau_n=\inf\{s\ge 0: |\exp(X_s)|\ge n\}\wedge n$. In this case, why is $\lim_{n\to \infty} \tau_n = \infty$ almost surely?

This means that almost surely, for any $n$, $|\exp(X_s)|\le n$ for $s\le n$, however, I cannot figure out why this must be the case.