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Here is a proof similar to what the OP has tried:

Let $\lambda$ be any eigenvalue of $AB$ with corresponding eigenvector $x$. Then

$$ABx = \lambda x \Rightarrow \\ BABx = B\lambda x \Rightarrow\\ BA(Bx) = \lambda (Bx) $$

which implies that $\lambda$ is an eigenvalue of $BA$ with a corresponding eigenvector $Bx$, provided $Bx$ is non-zero.

Suppose $Bx$ is the zero-vector. As $x$ is an eigenvector, it is non-zero, so $Bx = 0$ implies the $B$ is singular. Then both $AB$ and $BA$ are singular, and $\lambda = 0$ (as $ABx = 0$) is a common eigenvalue.

Thus, $AB$ and $BA$ have the same eigenvalues.

Here is a proof similar to what the OP has tried:

Let $\lambda$ be any eigenvalue of $AB$ with corresponding eigenvector $x$. Then

$$ABx = \lambda x \Rightarrow \\ BABx = B\lambda x \Rightarrow\\ BA(Bx) = \lambda (Bx) $$

which implies that $\lambda$ is an eigenvalue of $BA$ with a corresponding eigenvector $Bx$, provided $Bx$ is non-zero. If $Bx = 0$, then $ABx = 0$ implies that $\lambda = 0$.

Thus, $AB$ and $BA$ have the same non-zero eigenvalues.

Here is a proof similar to what the OP has tried:

Let $\lambda$ be any eigenvalue of $AB$ with corresponding eigenvector $x$. Then

$$ABx = \lambda x \Rightarrow \\ BABx = B\lambda x \Rightarrow\\ BA(Bx) = \lambda (Bx) $$

which implies that $\lambda$ is an eigenvalue of $BA$ with a corresponding eigenvector $Bx$, provided $Bx$ is non-zero. Suppose $Bx$ is the zero-vector. As $x$ is an eigenvector, it is non-zero, so $Bx = 0$ implies the $B$ is singular. Then both $AB$ and $BA$ are singular, and $\lambda = 0$ (as $ABx = 0$) is a common eigenvalue.

Thus, $AB$ and $BA$ have the same eigenvalues.

Here is a proof similar to what the OP has tried:

Let $\lambda$ be any eigenvalue of $AB$ with corresponding eigenvector $x$. Then

$$ABx = \lambda x \Rightarrow \\ BABx = B\lambda x \Rightarrow\\ BA(Bx) = \lambda (Bx) $$

which implies that $\lambda$ is an eigenvalue of $BA$ with a corresponding eigenvector $Bx$, provided $Bx$ is non-zero. 

Suppose $Bx$ is the zero-vector. As $x$ is an eigenvector, it is non-zero, so $Bx = 0$ implies the $B$ is singular. Then both $AB$ and $BA$ are singular, and $\lambda = 0$ (as $ABx = 0$) is a common eigenvalue.

Thus, $AB$ and $BA$ have the same eigenvalues.

Here is a proof similar to what the OP has tried:

Let $\lambda$ be any eigenvalue of $AB$ with corresponding eigenvector $x$. Then

$$ABx = \lambda \Rightarrow \\ BABx = B\lambda x = \lambda \Rightarrow\\ BA(Bx) = \lambda (Bx) $$

which implies that $\lambda$ is an eigenvalue of $BA$ with a corresponding eigenvector $Bx$, provided $Bx$ is non-zero. Suppose $Bx$ is the zero-vector. As $x$ is an eigenvector, it is non-zero, so $Bx = 0$ implies the $B$ is singular. Then both $AB$ and $BA$ are singular, and $\lambda = 0$ (as $ABx = 0$) is a common eigenvalue.

Thus, $AB$ and $BA$ have the same eigenvalues.

Here is a proof similar to what the OP has tried:

Let $\lambda$ be any eigenvalue of $AB$ with corresponding eigenvector $x$. Then

$$ABx = \lambda x \Rightarrow \\ BABx = B\lambda x \Rightarrow\\ BA(Bx) = \lambda (Bx) $$

which implies that $\lambda$ is an eigenvalue of $BA$ with a corresponding eigenvector $Bx$, provided $Bx$ is non-zero. Suppose $Bx$ is the zero-vector. As $x$ is an eigenvector, it is non-zero, so $Bx = 0$ implies the $B$ is singular. Then both $AB$ and $BA$ are singular, and $\lambda = 0$ (as $ABx = 0$) is a common eigenvalue.

Thus, $AB$ and $BA$ have the same eigenvalues.