The last digit of $3^3$ is 7.

The last digit of $3^{3^3}$ is 7.

The last digit of $3^{3^{3^3}}$ is 7.

The 3 statements above are all true. Is it true that for all integers $n\geq2$, $\underbrace{3^{3^{3^{3^{\cdot^{\cdot^3}}}}}}_{n\text{ number of 3's}}$ has a last digit of 7?

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