Find the solution of a differential equation in the form of a power series 
Find the solution of the differential equation $y''(x)=y(x)$ with $y(0)=1$ and $y'(0)=0 $ in the form of the power series $y(x)=\sum_{j=0}^{\infty}a_jx^j$
Use the following method of determining the coefficients $a_j$. Differentiate the power series twice term by term, then apply the identity theorem for power series to the series for y and y''

So differentiating I get:
$y'(x)=\sum_{j=0}^{\infty}(j+1)a_{j+1}x^j$ and $y''(x)=\sum_{j=0}^{\infty}(j+1)j\,a_{j+1}x^{j-1}$
Using the identity theorem then $y''(x)=y$ if $a_j=(j+1)j\,a_{j+1}$
And $y(0)=1=a_0$ and $y'(0)=0=a_1$
However I am unsure where to go from here? Thanks

The trick here is to compare two power series with equal powers in $x$. Following your results, we have
\begin{align}
 y(x) &= \sum_{j=0}^\infty a_j\,x^j,\\
 y''(x) &= \sum_{j=0}^\infty (j+1)j\,a_{j+1}\,x^{j-1}.
\end{align}
The original differential equation tells us that $y''(x) - y(x) = 0$, so in terms of power series, this would mean that
\begin{equation}
 \sum_{j=0}^\infty (j+1)j\,a_{j+1}\,x^{j-1} - \sum_{j=0}^\infty a_j\,x^j = \sum_{j=0}^\infty \Big[(j+1)j\,a_{j+1}\,x^{j-1} - a_j\,x^j\Big] = 0.
\end{equation}
We can't really conclude anything from this, unfortunately, because the two terms in the sum don't share a common $x$-factor (no pun intended), so we can't write it as
\begin{equation}
 \sum_{j=0}^\infty \Big[ \text{...something with $j$ and $a_j$...} \Big]\,x^j = 0,
\end{equation}
which would be very helpful.
The solution is to shift the summation index in the series for $y''(x)$, just as you did in the series for $y'(x)$. In other words, let's introduce $k = j-1$, i.e. $j = k+1$. Also --and this is important!--, we realise that the first term (for $j=0$) in the series for $y''(x)$ vanishes, since it evaluates to $0 \cdot (0+1) \cdot a_{0+1} = 0$. Therefore, we can drop that term and start the summation at $j=1$. So, we write
\begin{equation}
y''(x) = \sum_{j=0}^\infty (j+1)j\,a_{j+1}\,x^{j-1} = \sum_{j=1}^\infty (j+1)j\,a_{j+1}\,x^{j-1} = \sum_{k=0}^\infty (k+2)(k+1)\,a_{k+2}\,x^k.
\end{equation}
Note that the lower limit of the sum changes as the summation index is shifted. Now, we are able to write the differential equation as
\begin{align}
 y''(x) - y(x) &= \sum_{k=0}^\infty (k+2)(k+1)\,a_{k+2}\,x^k - \sum_{k=0}^\infty a_{k}\,x^k \\&= \sum_{k=0}^\infty \Big[(k+2)(k+1)\,a_{k+2}- a_k\Big]\,x^k = 0.
\end{align}
This must hold identically, for all $x$, so we obtain the following identity for the power series coefficients:
\begin{equation}
 (k+2)(k+1)\,a_{k+2}- a_k = 0.
\end{equation}
I'm sure you can take it from here, considering that the initial conditions give $a_0 = 1$ and $a_1 = 0$.
