Show that $e^{f(x)}$ is convex.

\begin{align} f(\lambda x + (1-\lambda)y) & \le \lambda f(x) + (1-\lambda) f(y) & & \text{because $f$ is convex.} \\[10pt] \text{Therefore } e^{f(\lambda x+(1-\lambda)y))} & \leq e^{\lambda f(x) + (1-\lambda) f(y)} & & \text{because $w\mapsto e^w$ is increasing,} \\[10pt] & = e^{\lambda v + (1-\lambda) w} \\[10pt] & \le \lambda e^v + (1-\lambda)e^w & & \text{because $w\mapsto e^w$ is convex, since} \\ & & & \text{its second derivative is positive,} \\[10pt] & = \lambda e^{f(x)} + (1-\lambda)e^{f(y)}. \end{align}

"From here, I want to bring the $\lambda$ and $1−λ$ terms down in front of the exponential, but I am stuck as to how to do this."

If you wish to continue from where you stuck: denoting $e^{f(x)}=A$, $e^{f(y)}=B$, you have to prove that $$ A^\lambda B^{1-\lambda}\le\lambda A+(1-\lambda)B. $$ Taking the logarithm on the both sides, it is equivalent to $$ \lambda\ln A+(1-\lambda)\ln B\le\ln(\lambda A+(1-\lambda)B) $$ which is the same as the definition of the logarithm being a concave function.