Showing $3^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}4^{n-k}}$. This is Exercise 1.9.3 of F. M. Goodman's "Algebra: Abstract and Concrete".

Show $$3^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}4^{n-k}}.$$

My Attempt:
I tried induction on $n$ as follows.
Base: Try when $n=1$. Then $LHS=3^1=3$ and
$$\begin{align}
RHS&=\sum_{k=0}^1{(-1)^k\binom{1}{k}4^{1-k}} \\
&=(-1)^0\binom{1}{0}4^1+(-1)^1\binom{1}{1}4^0 \\
&=4-1 \\
&=3
\end{align}$$ so the result holds for $n=1$.
Assume the result for $n=r\in\mathbb{N}$. Then $$3^r=\sum_{k=0}^r{(-1)^k\binom{r}{k}4^{r-k}}.$$
Consider when $n=r+1$: I have $3^{r+1}=3\cdot 3^r=3\sum_{k=0}^r{(-1)^k\binom{r}{k}4^{r-k}}$ and have considered
$$\sum_{k=0}^{r+1}{(-1)^k\binom{r+1}{k}4^{r+1-k}}$$ by writing $\binom{r+1}{k}=\binom{r}{k}+\binom{r}{k-1}$ but to no avail.

$$\begin{align}
\sum_{k=0}^n(-1)^k\binom nk 4^{n-k}
&=4^n\sum_{k=0}^n \binom nk \left(-\frac14\right)^k\\
&=4^n\left(1-\frac 14\right)^n\\
&=4^n\left(\frac 34\right)^n\\
&=3^n\end{align}$$
