Gambler's Ruin - Chance of winning increases by doubling the stakes

It is well-known that the probability of the gambler winning $N$ dollars before going bankrupt with starting fortune $k$ is 
\[P_k= \frac{1-(\frac{1-p}{p})^k}{1-(\frac{1-p}{p})^N},   p\ne q.\]

This is assuming that every time the gambler bets 1 dollar, and they need $N$ successes to win a total of $N$ dollars. 

Now if the bets are doubled, then the probability of winning $N$ dollars is the same as probability of winning $\frac{N}{2}$ dollars with one dollar bets with starting fortune of $\frac{k}{2}$. Thus the probability is 
\[P'_k=\frac{1-(\frac{1-p}{p})^{\frac{k}{2}}}{1-(\frac{1-p}{p})^{\frac{N}{2}}}.\]

It is enough to show that 
\[P'_k>P_k.\]
We have 
\[P_k= \frac{1-(\frac{1-p}{p})^k}{1-(\frac{1-p}{p})^N}=\frac{1-(\frac{1-p}{p})^{\frac{k}{2}}}{1-(\frac{1-p}{p})^{\frac{N}{2}}}\cdot \frac{1+(\frac{1-p}{p})^{\frac{k}{2}}}{1+(\frac{1-p}{p})^{\frac{N}{2}}}.\]
Hence
\[P_k=P'_k \big ( \frac{1+(\frac{1-p}{p})^{\frac{k}{2}}}{1+(\frac{1-p}{p})^{\frac{N}{2}}} \big).     (*)\]
Notice that since $p<\frac{1}{2}$, $$\frac{1-p}{p}>1,$$
thus $N>k$ implies 
\[1+(\frac{1-p}{p})^{\frac{N}{2}}>1+(\frac{1-p}{p})^{\frac{k}{2}}  \Rightarrow \frac{1+(\frac{1-p}{p})^{\frac{k}{2}}}{1+(\frac{1-p}{p})^{\frac{N}{2}}}<1. \]
In view of (*) this implies that 
\[P'_k >P_k.\]
The proof is complete!